charles law sample problem solving

Charles' Law Example Problem

Real-life applications for the ideal gas law at constant pressure

• Chemical Laws
• Periodic Table
• Projects & Experiments
• Scientific Method
• Biochemistry
• Physical Chemistry
• Medical Chemistry
• Chemistry In Everyday Life
• Famous Chemists
• Activities for Kids
• Abbreviations & Acronyms
• Weather & Climate
• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

Charles' law is a special case of the ideal gas law in which the pressure of a gas is constant. Charles' law states that volume is proportional to the absolute temperature of a gas at constant pressure. Doubling the temperature of gas doubles its volume, so long as the pressure and quantity of the gas are unchanged. 

This example problem shows how to use Charles' law to solve a gas law problem: A 600 mL sample of nitrogen is heated from 27 °C to 77 °C at constant pressure. What is the final volume?

The first step to solving gas law problems should be converting all temperatures to absolute temperatures . In other words, if the temperature is given in Celsius or Fahrenheit, convert it to Kelvin. (This is where the most commonplace mistakes are made in this type of homework problem.)

T K = 273 + °C T i = initial temperature = 27 °C T i K = 273 + 27 T i K = 300 K T f = final temperature = 77 °C T f K = 273 + 77 T f K = 350 K

The next step is to use Charles' law to find the final volume. Charles' law is expressed as:

V i /T i = V f /T f where V i and T i is the initial volume and temperature V f and T f is the final volume and temperature Solve the equation for V f : V f = V i T f /T i Enter the known values and solve for V f . V f = (600 mL)(350 K)/(300 K) V f = 700 mL Answer: The final volume after heating will be 700 mL.

More Examples of Charles' Law

If you think Charles' Law seems irrelevant to real-life situations, think again! By understanding the basics of the law, you'll know what to expect in a variety of real-world situations and once you know how to solve a problem using Charles' Law, you can make predictions and even start to plan new inventions. Here are several examples of situations in which Charles' Law is at play:

• If you take a basketball outside on a cold day, the ball shrinks a bit as the temperature is decreased. This is also the case with any inflated object and explains why it's a good idea to check your car's tire pressure when the temperature drops.
• If you over-inflate a pool float on a hot day, it can swell in the sun and burst.
• Pop-up turkey thermometers work based on Charles' law. As the turkey cooks, the gas inside the thermometer expands until it can "pop" the plunger.

Examples of Other Gas Laws

Charles' law is only one of the special cases of the ideal gas law that you may encounter. Each of the laws is named for the person who formulated it . It's good to know how to tell the gas laws apart and be able to cite examples of each one.

• Amonton's Law: Doubling temperature doubles pressure at constant volume and mass. Example: As automobile tires heat up when you drive, their pressure increases.
• Boyle's Law: Doubling pressure halves volume, at constant temperature and mass. Example: When you blow bubbles underwater, they expand as they rise to the surface.
• Avogadro's Law: Doubling the mass or number of moles of a gas doubles the volume at constant temperature and pressure. Example: Inhaling fills the lungs with air, expanding their volume.
• Examples of Solids, Liquids, and Gases
• What Is the Formula for Charles' Law?
• Gay-Lussac's Gas Law Examples
• The Formula for Boyle's Law
• Avogadro's Law Example Problem
• Gay-Lussac's Law Definition
• Charles's Law Definition in Chemistry
• Boyle's Law: Worked Chemistry Problems
• Gases - General Properties of Gases
• Boyle's Law Explained With Example Problem
• The Combined Gas Law in Chemistry
• Ideal Gas Law Example Problem
• How to Calculate Density of a Gas
• Temperature Conversions - Kelvin, Celsius, Fahrenheit
• Ideal Gas Law Test Questions
• What Is Avogadro's Law? Definition and Example

• Science Notes Posts
• Contact Science Notes
• Todd Helmenstine Biography
• Anne Helmenstine Biography
• Free Printable Periodic Tables (PDF and PNG)
• Periodic Table Wallpapers
• Interactive Periodic Table
• Periodic Table Posters
• How to Grow Crystals
• Chemistry Projects
• Fire and Flames Projects
• Holiday Science
• Chemistry Problems With Answers
• Physics Problems
• Unit Conversion Example Problems
• Chemistry Worksheets
• Biology Worksheets
• Periodic Table Worksheets
• Physical Science Worksheets
• Science Lab Worksheets
• My Amazon Books

Charles’s Law – Definition, Formula, Examples

Charles’s law or the law of volumes is an ideal gas law that states that the volume and temperature of a fixed amount of gas are proportional at constant pressure . Doubling the temperature of a gas doubles its volume. Halving the temperature of a gas halves its volume. The law takes its name from French scientist Jacques Charles, who formulated the law in the 1780s.

Charles’s law states that increasing the temperature of a gas at constant pressure increases its volume.

Charles’s Law Formula

There are a few ways to state Charles law as a formula:

V ∝ T V/T = k V = kT V 1 /T 1 = V 2 /T 2 V 2 /V 1 = T 2 /T 1 V 1 T 2 = V 2 T 1

Here, T is absolute temperature , V is volume, and k is a non-zero constant. Note that absolute temperature means Celsius and Fahrenheit temperature must be converted to Kelvin. The graph of volume versus pressure shows the linear relationship. Also, the line points toward the origin, although a gas could never reach it because it would change into a liquid or solid first.

Examples of Charles’s Law in Everyday Life

It’s easy to find examples of Charles’s law in everyday life.

• Hot air balloons fly based on Charles’s law. Heating the air in the balloon increases the balloon’s volume. This decreases its density, so the balloon rises in the air. To come down, chilling the air (not-heating-it) allows the balloon to deflate. The gas becomes more dense and the balloon sinks.
• If you take a filled balloon outside on a hot day, it expands (and may pop!). If you take it outdoors on a winter day, it deflates but returns to its normal volume when you take it indoors again. You can even use a balloon as a poor sort of thermometer, using Charles’s law.

Charles’s Law Example Calculation

A gas occupies 221 cm 3  at a temperature of 0 °C and pressure of 760 mm Hg. Find its volume at 100 °C.

First, don’t worry about the pressure. The number doesn’t enter into the calculation. All that matters is that it’s a constant.

Use the equation:

V 1 /T 1 = V 2 /T 2

Convert 0 °C and 100 °C to Kelvin:

V 1  = 221cm 3 ; T 1  = 273K (0 + 273); T 2  = 373K (100 + 273)

Plug the values into the equation and solve for V 2 :

V 1 /T 1  = V 2 /T 2 221cm 3  / 273K = V 2  / 373K V 2   = (221 cm 3 )(373K) / 273K V 2   = 302 cm 3

Find the final temperature of a sample of nitrogen gas at constant pressure if it starts at 27 °C and changes volume from 600 mL to 700 mL.

First convert the temperature to Kelvin.

T 1 = 273 + 27 T 1 = 300 K

Next, plug in the numbers.

V 1 /T 1  = V 2 /T 2 600 mL/300 K = 700 mL/T 2 (T 2 )(600 mL/300 K) = 700 mL T 2 = (700 mL)/(600 mL/300 K) T 2 = (700 mL)/(2mL/K) T 2 = 350 K

Why Temperature Must Be in Kelvin

Charles’s law calculations require temperature on an absolute scale, such as the Kelvin scale. So, using the formula requires converting from Celsius or Fahrenheit to Kelvin. There are two reasons for this. First, the negative temperatures on the Celsius and Fahrenheit scales could lead to impossible negative volume calculations. Second, the energy doesn’t scale properly using relative scales. So, a gas at 20 K has twice the energy of a gas at 10K, but the same is not true of as gas at 20 °C compared to 10 °C or 20 °F compared to 10 °F.

What Happens at Absolute Zero?

Like the other ideal gas laws, Charles’s law doesn’t apply under extreme conditions. It doesn’t make sense at absolute zero. First, matter can’t have zero volume. Second, a gas at constant pressure eventually changes into a liquid or solid as temperature drops.

• Fullick, P. (1994). Physics . Heinemann. ISBN 978-0-435-57078-1.
• Gay-Lussac, J. L. (1802). “Recherches sur la dilatation des gaz et des vapeurs” [ Research on the expansion of gases and vapors ]. Annales de Chimie . 43: 137–75.
• Krönig, A. (1856). “ Grundzüge einer Theorie der Gase “. Annalen der Physik . 99 (10): 315–22. doi:10.1002/andp.18561751008

Related Posts

Charles' Law Calculator

Table of contents

The Charles' law calculator is a simple tool that describes the basic parameters of an ideal gas in an isobaric process . In the text, you can find the answer to the question "What is Charles' law?", learn what the Charles' law formula looks like, and read how to solve thermodynamic problems with some Charles' law examples.

In case you need to work out the results for an isochoric process, check our Gay-Lussac's law calculator .

Charles' law definition

Charles' law (sometimes referred to as the law of volumes) describes the relationship between the volume of a gas and its temperature when the pressure and the mass of the gas are constant . It states that the volume is proportional to the absolute temperature .

There are a few other ways we can write the Charles' law definition, one of which is: the ratio of the volume and the temperature of the gas in a closed system is constant as long as the pressure is unchanged.

Charles' law describes the behavior of an ideal gas (gases that we can characterize by the ideal gas law equation ) during an isobaric process , which means that the pressure remains constant during the transition.

Charles' law formula

Based on the definition of Charles' law, we can write the Charles' law equation in the following way:

V₁ / T₁ = V₂ / T₂ ,

where V₁ and T₁ are the initial volume and temperature, respectively. Similarly, V₂ and T₂ are the final values of these gas parameters.

How does this Charles' law calculator work? First, you need to insert three of the parameters, and the fourth is automatically calculated for you. Let's say we want to find the final volume, then the Charles' law formula yields:

V₂ = V₁ / T₁ × T₂ .

If you prefer to set the final volume and want to estimate the resulting temperature, then the equation of Charles' law changes to:

T₂ = T₁ / V₁ × V₂ .

In the last group of variables, you can also define the pressure and see how many moles of atoms or molecules there are in a container.

💡 If the temperature is constant during the transition, it's an isothermal process. In such a case, you can quickly estimate its parameters with Omni's Boyle's law calculator !

Charles' law examples

We can use Charles' law calculator to solve some thermodynamic problems. Let's see how it works:

Imagine that we have a ball pumped full of air. Its initial volume is equal to 2 liters , and it lies on a beach where the temperature is 35 °C . We then move it to an air-conditioned room with a temperature of 15 °C . How does the volume of the ball change?

First of all, the Charles' law formula requires the absolute values of temperatures so we have to convert them into Kelvin:

T₁ = 35 °C = 308.15 K , T₂ = 15 °C = 288.15 K .

Then we can apply the Charles' law equation in the form where the final volume is being evaluated:

V₂ = V₁ / T₁ × T₂ = 2 l / 308.15 K × 288.15 K = 1.8702 l .

We can see that the volume decreases when we move the ball from a warmer to a cooler place . Sometimes you can experience that effect while changing your location or simply leaving an object alone when the weather turns. The ball seems under-inflated, and somebody may think there is a hole, causing the air to leak. Fortunately, it's only physics, so you don't have to buy another ball – just inflate the one you have and enjoy!

One tiny remark – air is an example of a real gas, so the outcome is only an approximation , but as long as we avoid extreme conditions (pressure, temperature). The result is sufficiently close to the actual value.

In the second problem, we heat an easily-stretched container. It's filled with nitrogen, which is a good approximation of an ideal gas. We can find that its initial volume is 0.03 ft³ at room temperature, 295 K . Then we put it close to the heating source and leave it for a while. After a few minutes, its volume has increased to 0.062 ft³ . With all of this data, can we estimate the temperature of our heater?

Let's apply the Charles' law formula and rewrite it in a form so that we can work out the temperature:

T₂ = T₁ / V₁ × V₂ = 295 K / 0.03 ft³ × 0.062 ft³ = 609.7 K .

We can write the outcome in the more amiable form T₂ = 336.5 °C or T₂ = 637.7 °F .

This is a great example that shows us that we can use this kind of device as a thermometer ! Well, it's not a very practical method and is probably not as precise as the common ones, but it still makes you think, what other unusual applications can you get from other everyday objects?

What is Charles' law application in real life?

There are actually various areas where we can use Charles' law. Here is a list of a few of the most popular and intriguing examples:

Balloon flight – You must have seen a balloon in the sky at least once in your life. Have you ever wondered how it is possible for it to fly and why they are equipped with fire or other heating sources on board? Charles' law is the answer! Whenever the air is heated, its volume increases . As a result, the same amount (mass) of gas occupies a greater space, which means the density decreases. The buoyancy of the surrounding air does the rest of the job, so the balloon begins to float.

The steering at any given direction is probably a different story, but we can explain the general concept of the up and down movement with Charles' law .

Liquid nitrogen experiments – Have you ever seen an experiment where someone puts a ball or balloon inside a container filled with liquid nitrogen and then moves outside? Firstly, it shrinks no matter how big it is at the beginning. Then, after it is freed, it returns to its initial state. Once again, whenever the temperature changes, so does the volume.

Thermometer – As shown in the previous section, it is possible to construct a device that measures temperature based on Charles' law. Although we must be aware of its limitations, which are basically the object's tensile strength and resistance to high temperatures, we can invent an original device that works perfectly to suit our needs. Whenever you are uncertain about the outcome, check this Charles' law calculator to find the answer.

Other thermodynamic processes

Charles' law, Boyle's law, and Gay-Lussac's law are among the fundamental laws which describe the vast majority of thermodynamic processes. We have gathered all of the basic gas transitions in our combined gas law calculator , where you can evaluate not only the final temperature, pressure, or volume but also the internal energy change or work done by gas.

What is a Charles' law?

Charles's law states that the volume (V) of a gas is directly proportional to the temperature (T) when pressure is kept constant . The temperature must be measured with the Kelvin scale . When we compare the substance under initial ( V₁ , T₁ ) and final conditions ( V₂ , T₂ ), you can write Charles' law as V₁/T₁ = V₂/T₂ . As the temperature increases , the volume of gas also increases proportionally.

When was Charles' law discovered?

The law was experimentally determined by ballooning pioneer Jacques Charles in 1787 . Unfortunately, Charles never published the work for which he is remembered. Similar studies had been carried out 100 years earlier by Guillaume Amontons , and in 1808 Joseph Gay-Lussac made the final measurements and published generalized results for gases .

How can I find T₂ in Charles' law equation?

Let's say the volume of gas was compressed from 3 liters to 2 liters at an initial temperature of 25°C . To find T₂ (final temperature) in Charles' law :

• Convert initial temperature T₁ to the Kelvins: T₁ + 273.15 = 298.15 K.
• Solve Charles' law for T₂ : T₂ = (T₁×V₂)/V₁.
• Enter data : T₂ = (298.15 K×2 L)/3 L = 198.77 K.

What is the initial volume if gas was heated from 270°C to 342°C?

662.2 mL , assuming that the final volume is 750 mL . To calculate this:

• Convert temperatures to Kelvins: T₁ = 543.2 K, T₂ = 615.2 K.
• Write Charles' law : V₁/T₁ = V₂/T₂.
• Solve for V₁ : V₁ = (T₁×V₂)/T₂.
• Substitute values : V₁ = (543.2 K×750 mL)/615.2 K = 662.2 mL.

What are the limitations of Charles' law?

Charles' law can only be applied to ideal gases (gases whose molecules neither attract nor repel each othert ). Charles' law is valid for real gases only in the high-temperature range and at low pressures . Note that the relationship between volume and temperature is not linear at high pressures .

.css-slt4t3.css-slt4t3{color:#2B3148;background-color:transparent;font-family:"Roboto","Helvetica","Arial",sans-serif;font-size:20px;line-height:24px;overflow:visible;padding-top:0px;position:relative;}.css-slt4t3.css-slt4t3:after{content:'';-webkit-transform:scale(0);-moz-transform:scale(0);-ms-transform:scale(0);transform:scale(0);position:absolute;border:2px solid #EA9430;border-radius:2px;inset:-8px;z-index:1;}.css-slt4t3 .js-external-link-button.link-like,.css-slt4t3 .js-external-link-anchor{color:inherit;border-radius:1px;-webkit-text-decoration:underline;text-decoration:underline;}.css-slt4t3 .js-external-link-button.link-like:hover,.css-slt4t3 .js-external-link-anchor:hover,.css-slt4t3 .js-external-link-button.link-like:active,.css-slt4t3 .js-external-link-anchor:active{text-decoration-thickness:2px;text-shadow:1px 0 0;}.css-slt4t3 .js-external-link-button.link-like:focus-visible,.css-slt4t3 .js-external-link-anchor:focus-visible{outline:transparent 2px dotted;box-shadow:0 0 0 2px #6314E6;}.css-slt4t3 p,.css-slt4t3 div{margin:0px;display:block;}.css-slt4t3 pre{margin:0px;display:block;}.css-slt4t3 pre code{display:block;width:-webkit-fit-content;width:-moz-fit-content;width:fit-content;}.css-slt4t3 pre:not(:first-child){padding-top:8px;}.css-slt4t3 ul,.css-slt4t3 ol{display:block margin:0px;padding-left:20px;}.css-slt4t3 ul li,.css-slt4t3 ol li{padding-top:8px;}.css-slt4t3 ul ul,.css-slt4t3 ol ul,.css-slt4t3 ul ol,.css-slt4t3 ol ol{padding-top:0px;}.css-slt4t3 ul:not(:first-child),.css-slt4t3 ol:not(:first-child){padding-top:4px;} .css-4okk7a{margin:auto;background-color:white;overflow:auto;overflow-wrap:break-word;word-break:break-word;}.css-4okk7a code,.css-4okk7a kbd,.css-4okk7a pre,.css-4okk7a samp{font-family:monospace;}.css-4okk7a code{padding:2px 4px;color:#444;background:#ddd;border-radius:4px;}.css-4okk7a figcaption,.css-4okk7a caption{text-align:center;}.css-4okk7a figcaption{font-size:12px;font-style:italic;overflow:hidden;}.css-4okk7a h3{font-size:1.75rem;}.css-4okk7a h4{font-size:1.5rem;}.css-4okk7a .mathBlock{font-size:24px;-webkit-padding-start:4px;padding-inline-start:4px;}.css-4okk7a .mathBlock .katex{font-size:24px;text-align:left;}.css-4okk7a .math-inline{background-color:#f0f0f0;display:inline-block;font-size:inherit;padding:0 3px;}.css-4okk7a .videoBlock,.css-4okk7a .imageBlock{margin-bottom:16px;}.css-4okk7a .imageBlock__image-align--left,.css-4okk7a .videoBlock__video-align--left{float:left;}.css-4okk7a .imageBlock__image-align--right,.css-4okk7a .videoBlock__video-align--right{float:right;}.css-4okk7a .imageBlock__image-align--center,.css-4okk7a .videoBlock__video-align--center{display:block;margin-left:auto;margin-right:auto;clear:both;}.css-4okk7a .imageBlock__image-align--none,.css-4okk7a .videoBlock__video-align--none{clear:both;margin-left:0;margin-right:0;}.css-4okk7a .videoBlock__video--wrapper{position:relative;padding-bottom:56.25%;height:0;}.css-4okk7a .videoBlock__video--wrapper iframe{position:absolute;top:0;left:0;width:100%;height:100%;}.css-4okk7a .videoBlock__caption{text-align:left;}@font-face{font-family:'KaTeX_AMS';src:url(/katex-fonts/KaTeX_AMS-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_AMS-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_AMS-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Caligraphic';src:url(/katex-fonts/KaTeX_Caligraphic-Bold.woff2) format('woff2'),url(/katex-fonts/KaTeX_Caligraphic-Bold.woff) format('woff'),url(/katex-fonts/KaTeX_Caligraphic-Bold.ttf) format('truetype');font-weight:bold;font-style:normal;}@font-face{font-family:'KaTeX_Caligraphic';src:url(/katex-fonts/KaTeX_Caligraphic-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Caligraphic-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Caligraphic-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Fraktur';src:url(/katex-fonts/KaTeX_Fraktur-Bold.woff2) format('woff2'),url(/katex-fonts/KaTeX_Fraktur-Bold.woff) format('woff'),url(/katex-fonts/KaTeX_Fraktur-Bold.ttf) format('truetype');font-weight:bold;font-style:normal;}@font-face{font-family:'KaTeX_Fraktur';src:url(/katex-fonts/KaTeX_Fraktur-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Fraktur-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Fraktur-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Main';src:url(/katex-fonts/KaTeX_Main-Bold.woff2) format('woff2'),url(/katex-fonts/KaTeX_Main-Bold.woff) format('woff'),url(/katex-fonts/KaTeX_Main-Bold.ttf) format('truetype');font-weight:bold;font-style:normal;}@font-face{font-family:'KaTeX_Main';src:url(/katex-fonts/KaTeX_Main-BoldItalic.woff2) format('woff2'),url(/katex-fonts/KaTeX_Main-BoldItalic.woff) format('woff'),url(/katex-fonts/KaTeX_Main-BoldItalic.ttf) format('truetype');font-weight:bold;font-style:italic;}@font-face{font-family:'KaTeX_Main';src:url(/katex-fonts/KaTeX_Main-Italic.woff2) format('woff2'),url(/katex-fonts/KaTeX_Main-Italic.woff) format('woff'),url(/katex-fonts/KaTeX_Main-Italic.ttf) format('truetype');font-weight:normal;font-style:italic;}@font-face{font-family:'KaTeX_Main';src:url(/katex-fonts/KaTeX_Main-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Main-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Main-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Math';src:url(/katex-fonts/KaTeX_Math-BoldItalic.woff2) format('woff2'),url(/katex-fonts/KaTeX_Math-BoldItalic.woff) format('woff'),url(/katex-fonts/KaTeX_Math-BoldItalic.ttf) format('truetype');font-weight:bold;font-style:italic;}@font-face{font-family:'KaTeX_Math';src:url(/katex-fonts/KaTeX_Math-Italic.woff2) format('woff2'),url(/katex-fonts/KaTeX_Math-Italic.woff) format('woff'),url(/katex-fonts/KaTeX_Math-Italic.ttf) format('truetype');font-weight:normal;font-style:italic;}@font-face{font-family:'KaTeX_SansSerif';src:url(/katex-fonts/KaTeX_SansSerif-Bold.woff2) format('woff2'),url(/katex-fonts/KaTeX_SansSerif-Bold.woff) format('woff'),url(/katex-fonts/KaTeX_SansSerif-Bold.ttf) format('truetype');font-weight:bold;font-style:normal;}@font-face{font-family:'KaTeX_SansSerif';src:url(/katex-fonts/KaTeX_SansSerif-Italic.woff2) format('woff2'),url(/katex-fonts/KaTeX_SansSerif-Italic.woff) format('woff'),url(/katex-fonts/KaTeX_SansSerif-Italic.ttf) format('truetype');font-weight:normal;font-style:italic;}@font-face{font-family:'KaTeX_SansSerif';src:url(/katex-fonts/KaTeX_SansSerif-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_SansSerif-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_SansSerif-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Script';src:url(/katex-fonts/KaTeX_Script-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Script-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Script-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Size1';src:url(/katex-fonts/KaTeX_Size1-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Size1-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Size1-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Size2';src:url(/katex-fonts/KaTeX_Size2-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Size2-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Size2-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Size3';src:url(/katex-fonts/KaTeX_Size3-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Size3-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Size3-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Size4';src:url(/katex-fonts/KaTeX_Size4-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Size4-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Size4-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}@font-face{font-family:'KaTeX_Typewriter';src:url(/katex-fonts/KaTeX_Typewriter-Regular.woff2) format('woff2'),url(/katex-fonts/KaTeX_Typewriter-Regular.woff) format('woff'),url(/katex-fonts/KaTeX_Typewriter-Regular.ttf) format('truetype');font-weight:normal;font-style:normal;}.css-4okk7a .katex{font:normal 1.21em KaTeX_Main,Times New Roman,serif;line-height:1.2;text-indent:0;text-rendering:auto;}.css-4okk7a .katex *{-ms-high-contrast-adjust:none!important;border-color:currentColor;}.css-4okk7a .katex .katex-version::after{content:'0.13.13';}.css-4okk7a .katex .katex-mathml{position:absolute;clip:rect(1px, 1px, 1px, 1px);padding:0;border:0;height:1px;width:1px;overflow:hidden;}.css-4okk7a .katex .katex-html>.newline{display:block;}.css-4okk7a .katex .base{position:relative;display:inline-block;white-space:nowrap;width:-webkit-min-content;width:-moz-min-content;width:-webkit-min-content;width:-moz-min-content;width:min-content;}.css-4okk7a .katex .strut{display:inline-block;}.css-4okk7a .katex .textbf{font-weight:bold;}.css-4okk7a .katex .textit{font-style:italic;}.css-4okk7a .katex .textrm{font-family:KaTeX_Main;}.css-4okk7a .katex .textsf{font-family:KaTeX_SansSerif;}.css-4okk7a .katex .texttt{font-family:KaTeX_Typewriter;}.css-4okk7a .katex .mathnormal{font-family:KaTeX_Math;font-style:italic;}.css-4okk7a .katex .mathit{font-family:KaTeX_Main;font-style:italic;}.css-4okk7a .katex .mathrm{font-style:normal;}.css-4okk7a .katex .mathbf{font-family:KaTeX_Main;font-weight:bold;}.css-4okk7a .katex .boldsymbol{font-family:KaTeX_Math;font-weight:bold;font-style:italic;}.css-4okk7a .katex .amsrm{font-family:KaTeX_AMS;}.css-4okk7a .katex .mathbb,.css-4okk7a .katex .textbb{font-family:KaTeX_AMS;}.css-4okk7a .katex .mathcal{font-family:KaTeX_Caligraphic;}.css-4okk7a .katex .mathfrak,.css-4okk7a .katex .textfrak{font-family:KaTeX_Fraktur;}.css-4okk7a .katex .mathtt{font-family:KaTeX_Typewriter;}.css-4okk7a .katex .mathscr,.css-4okk7a .katex .textscr{font-family:KaTeX_Script;}.css-4okk7a .katex .mathsf,.css-4okk7a .katex .textsf{font-family:KaTeX_SansSerif;}.css-4okk7a .katex .mathboldsf,.css-4okk7a .katex .textboldsf{font-family:KaTeX_SansSerif;font-weight:bold;}.css-4okk7a .katex .mathitsf,.css-4okk7a .katex .textitsf{font-family:KaTeX_SansSerif;font-style:italic;}.css-4okk7a .katex .mainrm{font-family:KaTeX_Main;font-style:normal;}.css-4okk7a .katex .vlist-t{display:inline-table;table-layout:fixed;border-collapse:collapse;}.css-4okk7a .katex .vlist-r{display:table-row;}.css-4okk7a .katex .vlist{display:table-cell;vertical-align:bottom;position:relative;}.css-4okk7a .katex .vlist>span{display:block;height:0;position:relative;}.css-4okk7a .katex .vlist>span>span{display:inline-block;}.css-4okk7a .katex .vlist>span>.pstrut{overflow:hidden;width:0;}.css-4okk7a .katex .vlist-t2{margin-right:-2px;}.css-4okk7a .katex .vlist-s{display:table-cell;vertical-align:bottom;font-size:1px;width:2px;min-width:2px;}.css-4okk7a .katex .vbox{display:-webkit-inline-box;display:-webkit-inline-flex;display:-ms-inline-flexbox;display:inline-flex;-webkit-flex-direction:column;-ms-flex-direction:column;flex-direction:column;-webkit-align-items:baseline;-webkit-box-align:baseline;-ms-flex-align:baseline;align-items:baseline;}.css-4okk7a .katex .hbox{display:-webkit-inline-box;display:-webkit-inline-flex;display:-ms-inline-flexbox;display:inline-flex;-webkit-flex-direction:row;-ms-flex-direction:row;flex-direction:row;width:100%;}.css-4okk7a .katex .thinbox{display:-webkit-inline-box;display:-webkit-inline-flex;display:-ms-inline-flexbox;display:inline-flex;-webkit-flex-direction:row;-ms-flex-direction:row;flex-direction:row;width:0;max-width:0;}.css-4okk7a .katex .msupsub{text-align:left;}.css-4okk7a .katex .mfrac>span>span{text-align:center;}.css-4okk7a .katex .mfrac .frac-line{display:inline-block;width:100%;border-bottom-style:solid;}.css-4okk7a .katex .mfrac .frac-line,.css-4okk7a .katex .overline .overline-line,.css-4okk7a .katex .underline .underline-line,.css-4okk7a .katex .hline,.css-4okk7a .katex .hdashline,.css-4okk7a .katex .rule{min-height:1px;}.css-4okk7a .katex .mspace{display:inline-block;}.css-4okk7a .katex .llap,.css-4okk7a .katex .rlap,.css-4okk7a .katex .clap{width:0;position:relative;}.css-4okk7a .katex .llap>.inner,.css-4okk7a .katex .rlap>.inner,.css-4okk7a .katex .clap>.inner{position:absolute;}.css-4okk7a .katex .llap>.fix,.css-4okk7a .katex .rlap>.fix,.css-4okk7a .katex .clap>.fix{display:inline-block;}.css-4okk7a .katex .llap>.inner{right:0;}.css-4okk7a .katex .rlap>.inner,.css-4okk7a .katex .clap>.inner{left:0;}.css-4okk7a .katex .clap>.inner>span{margin-left:-50%;margin-right:50%;}.css-4okk7a .katex .rule{display:inline-block;border:solid 0;position:relative;}.css-4okk7a .katex .overline .overline-line,.css-4okk7a .katex .underline .underline-line,.css-4okk7a .katex .hline{display:inline-block;width:100%;border-bottom-style:solid;}.css-4okk7a .katex .hdashline{display:inline-block;width:100%;border-bottom-style:dashed;}.css-4okk7a .katex .sqrt>.root{margin-left:0.27777778em;margin-right:-0.55555556em;}.css-4okk7a .katex .sizing.reset-size1.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size1{font-size:1em;}.css-4okk7a .katex .sizing.reset-size1.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size2{font-size:1.2em;}.css-4okk7a .katex .sizing.reset-size1.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size3{font-size:1.4em;}.css-4okk7a .katex .sizing.reset-size1.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size4{font-size:1.6em;}.css-4okk7a .katex .sizing.reset-size1.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size5{font-size:1.8em;}.css-4okk7a .katex .sizing.reset-size1.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size6{font-size:2em;}.css-4okk7a .katex .sizing.reset-size1.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size7{font-size:2.4em;}.css-4okk7a .katex .sizing.reset-size1.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size8{font-size:2.88em;}.css-4okk7a .katex .sizing.reset-size1.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size9{font-size:3.456em;}.css-4okk7a .katex .sizing.reset-size1.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size10{font-size:4.148em;}.css-4okk7a .katex .sizing.reset-size1.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size1.size11{font-size:4.976em;}.css-4okk7a .katex .sizing.reset-size2.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size1{font-size:0.83333333em;}.css-4okk7a .katex .sizing.reset-size2.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size2{font-size:1em;}.css-4okk7a .katex .sizing.reset-size2.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size3{font-size:1.16666667em;}.css-4okk7a .katex .sizing.reset-size2.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size4{font-size:1.33333333em;}.css-4okk7a .katex .sizing.reset-size2.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size5{font-size:1.5em;}.css-4okk7a .katex .sizing.reset-size2.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size6{font-size:1.66666667em;}.css-4okk7a .katex .sizing.reset-size2.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size7{font-size:2em;}.css-4okk7a .katex .sizing.reset-size2.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size8{font-size:2.4em;}.css-4okk7a .katex .sizing.reset-size2.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size9{font-size:2.88em;}.css-4okk7a .katex .sizing.reset-size2.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size10{font-size:3.45666667em;}.css-4okk7a .katex .sizing.reset-size2.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size2.size11{font-size:4.14666667em;}.css-4okk7a .katex .sizing.reset-size3.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size1{font-size:0.71428571em;}.css-4okk7a .katex .sizing.reset-size3.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size2{font-size:0.85714286em;}.css-4okk7a .katex .sizing.reset-size3.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size3{font-size:1em;}.css-4okk7a .katex .sizing.reset-size3.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size4{font-size:1.14285714em;}.css-4okk7a .katex .sizing.reset-size3.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size5{font-size:1.28571429em;}.css-4okk7a .katex .sizing.reset-size3.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size6{font-size:1.42857143em;}.css-4okk7a .katex .sizing.reset-size3.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size7{font-size:1.71428571em;}.css-4okk7a .katex .sizing.reset-size3.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size8{font-size:2.05714286em;}.css-4okk7a .katex .sizing.reset-size3.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size9{font-size:2.46857143em;}.css-4okk7a .katex .sizing.reset-size3.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size10{font-size:2.96285714em;}.css-4okk7a .katex .sizing.reset-size3.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size3.size11{font-size:3.55428571em;}.css-4okk7a .katex .sizing.reset-size4.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size1{font-size:0.625em;}.css-4okk7a .katex .sizing.reset-size4.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size2{font-size:0.75em;}.css-4okk7a .katex .sizing.reset-size4.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size3{font-size:0.875em;}.css-4okk7a .katex .sizing.reset-size4.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size4{font-size:1em;}.css-4okk7a .katex .sizing.reset-size4.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size5{font-size:1.125em;}.css-4okk7a .katex .sizing.reset-size4.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size6{font-size:1.25em;}.css-4okk7a .katex .sizing.reset-size4.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size7{font-size:1.5em;}.css-4okk7a .katex .sizing.reset-size4.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size8{font-size:1.8em;}.css-4okk7a .katex .sizing.reset-size4.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size9{font-size:2.16em;}.css-4okk7a .katex .sizing.reset-size4.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size10{font-size:2.5925em;}.css-4okk7a .katex .sizing.reset-size4.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size4.size11{font-size:3.11em;}.css-4okk7a .katex .sizing.reset-size5.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size1{font-size:0.55555556em;}.css-4okk7a .katex .sizing.reset-size5.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size2{font-size:0.66666667em;}.css-4okk7a .katex .sizing.reset-size5.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size3{font-size:0.77777778em;}.css-4okk7a .katex .sizing.reset-size5.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size4{font-size:0.88888889em;}.css-4okk7a .katex .sizing.reset-size5.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size5{font-size:1em;}.css-4okk7a .katex .sizing.reset-size5.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size6{font-size:1.11111111em;}.css-4okk7a .katex .sizing.reset-size5.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size7{font-size:1.33333333em;}.css-4okk7a .katex .sizing.reset-size5.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size8{font-size:1.6em;}.css-4okk7a .katex .sizing.reset-size5.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size9{font-size:1.92em;}.css-4okk7a .katex .sizing.reset-size5.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size10{font-size:2.30444444em;}.css-4okk7a .katex .sizing.reset-size5.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size5.size11{font-size:2.76444444em;}.css-4okk7a .katex .sizing.reset-size6.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size1{font-size:0.5em;}.css-4okk7a .katex .sizing.reset-size6.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size2{font-size:0.6em;}.css-4okk7a .katex .sizing.reset-size6.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size3{font-size:0.7em;}.css-4okk7a .katex .sizing.reset-size6.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size4{font-size:0.8em;}.css-4okk7a .katex .sizing.reset-size6.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size5{font-size:0.9em;}.css-4okk7a .katex .sizing.reset-size6.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size6{font-size:1em;}.css-4okk7a .katex .sizing.reset-size6.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size7{font-size:1.2em;}.css-4okk7a .katex .sizing.reset-size6.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size8{font-size:1.44em;}.css-4okk7a .katex .sizing.reset-size6.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size9{font-size:1.728em;}.css-4okk7a .katex .sizing.reset-size6.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size10{font-size:2.074em;}.css-4okk7a .katex .sizing.reset-size6.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size6.size11{font-size:2.488em;}.css-4okk7a .katex .sizing.reset-size7.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size1{font-size:0.41666667em;}.css-4okk7a .katex .sizing.reset-size7.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size2{font-size:0.5em;}.css-4okk7a .katex .sizing.reset-size7.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size3{font-size:0.58333333em;}.css-4okk7a .katex .sizing.reset-size7.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size4{font-size:0.66666667em;}.css-4okk7a .katex .sizing.reset-size7.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size5{font-size:0.75em;}.css-4okk7a .katex .sizing.reset-size7.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size6{font-size:0.83333333em;}.css-4okk7a .katex .sizing.reset-size7.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size7{font-size:1em;}.css-4okk7a .katex .sizing.reset-size7.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size8{font-size:1.2em;}.css-4okk7a .katex .sizing.reset-size7.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size9{font-size:1.44em;}.css-4okk7a .katex .sizing.reset-size7.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size10{font-size:1.72833333em;}.css-4okk7a .katex .sizing.reset-size7.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size7.size11{font-size:2.07333333em;}.css-4okk7a .katex .sizing.reset-size8.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size1{font-size:0.34722222em;}.css-4okk7a .katex .sizing.reset-size8.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size2{font-size:0.41666667em;}.css-4okk7a .katex .sizing.reset-size8.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size3{font-size:0.48611111em;}.css-4okk7a .katex .sizing.reset-size8.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size4{font-size:0.55555556em;}.css-4okk7a .katex .sizing.reset-size8.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size5{font-size:0.625em;}.css-4okk7a .katex .sizing.reset-size8.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size6{font-size:0.69444444em;}.css-4okk7a .katex .sizing.reset-size8.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size7{font-size:0.83333333em;}.css-4okk7a .katex .sizing.reset-size8.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size8{font-size:1em;}.css-4okk7a .katex .sizing.reset-size8.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size9{font-size:1.2em;}.css-4okk7a .katex .sizing.reset-size8.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size10{font-size:1.44027778em;}.css-4okk7a .katex .sizing.reset-size8.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size8.size11{font-size:1.72777778em;}.css-4okk7a .katex .sizing.reset-size9.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size1{font-size:0.28935185em;}.css-4okk7a .katex .sizing.reset-size9.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size2{font-size:0.34722222em;}.css-4okk7a .katex .sizing.reset-size9.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size3{font-size:0.40509259em;}.css-4okk7a .katex .sizing.reset-size9.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size4{font-size:0.46296296em;}.css-4okk7a .katex .sizing.reset-size9.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size5{font-size:0.52083333em;}.css-4okk7a .katex .sizing.reset-size9.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size6{font-size:0.5787037em;}.css-4okk7a .katex .sizing.reset-size9.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size7{font-size:0.69444444em;}.css-4okk7a .katex .sizing.reset-size9.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size8{font-size:0.83333333em;}.css-4okk7a .katex .sizing.reset-size9.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size9{font-size:1em;}.css-4okk7a .katex .sizing.reset-size9.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size10{font-size:1.20023148em;}.css-4okk7a .katex .sizing.reset-size9.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size9.size11{font-size:1.43981481em;}.css-4okk7a .katex .sizing.reset-size10.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size1{font-size:0.24108004em;}.css-4okk7a .katex .sizing.reset-size10.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size2{font-size:0.28929605em;}.css-4okk7a .katex .sizing.reset-size10.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size3{font-size:0.33751205em;}.css-4okk7a .katex .sizing.reset-size10.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size4{font-size:0.38572806em;}.css-4okk7a .katex .sizing.reset-size10.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size5{font-size:0.43394407em;}.css-4okk7a .katex .sizing.reset-size10.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size6{font-size:0.48216008em;}.css-4okk7a .katex .sizing.reset-size10.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size7{font-size:0.57859209em;}.css-4okk7a .katex .sizing.reset-size10.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size8{font-size:0.69431051em;}.css-4okk7a .katex .sizing.reset-size10.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size9{font-size:0.83317261em;}.css-4okk7a .katex .sizing.reset-size10.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size10{font-size:1em;}.css-4okk7a .katex .sizing.reset-size10.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size10.size11{font-size:1.19961427em;}.css-4okk7a .katex .sizing.reset-size11.size1,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size1{font-size:0.20096463em;}.css-4okk7a .katex .sizing.reset-size11.size2,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size2{font-size:0.24115756em;}.css-4okk7a .katex .sizing.reset-size11.size3,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size3{font-size:0.28135048em;}.css-4okk7a .katex .sizing.reset-size11.size4,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size4{font-size:0.32154341em;}.css-4okk7a .katex .sizing.reset-size11.size5,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size5{font-size:0.36173633em;}.css-4okk7a .katex .sizing.reset-size11.size6,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size6{font-size:0.40192926em;}.css-4okk7a .katex .sizing.reset-size11.size7,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size7{font-size:0.48231511em;}.css-4okk7a .katex .sizing.reset-size11.size8,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size8{font-size:0.57877814em;}.css-4okk7a .katex .sizing.reset-size11.size9,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size9{font-size:0.69453376em;}.css-4okk7a .katex .sizing.reset-size11.size10,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size10{font-size:0.83360129em;}.css-4okk7a .katex .sizing.reset-size11.size11,.css-4okk7a .katex .fontsize-ensurer.reset-size11.size11{font-size:1em;}.css-4okk7a .katex .delimsizing.size1{font-family:KaTeX_Size1;}.css-4okk7a .katex .delimsizing.size2{font-family:KaTeX_Size2;}.css-4okk7a .katex .delimsizing.size3{font-family:KaTeX_Size3;}.css-4okk7a .katex .delimsizing.size4{font-family:KaTeX_Size4;}.css-4okk7a .katex .delimsizing.mult .delim-size1>span{font-family:KaTeX_Size1;}.css-4okk7a .katex .delimsizing.mult .delim-size4>span{font-family:KaTeX_Size4;}.css-4okk7a .katex .nulldelimiter{display:inline-block;width:0.12em;}.css-4okk7a .katex .delimcenter{position:relative;}.css-4okk7a .katex .op-symbol{position:relative;}.css-4okk7a .katex .op-symbol.small-op{font-family:KaTeX_Size1;}.css-4okk7a .katex .op-symbol.large-op{font-family:KaTeX_Size2;}.css-4okk7a .katex .op-limits>.vlist-t{text-align:center;}.css-4okk7a .katex .accent>.vlist-t{text-align:center;}.css-4okk7a .katex .accent .accent-body{position:relative;}.css-4okk7a .katex .accent .accent-body:not(.accent-full){width:0;}.css-4okk7a .katex .overlay{display:block;}.css-4okk7a .katex .mtable .vertical-separator{display:inline-block;min-width:1px;}.css-4okk7a .katex .mtable .arraycolsep{display:inline-block;}.css-4okk7a .katex .mtable .col-align-c>.vlist-t{text-align:center;}.css-4okk7a .katex .mtable .col-align-l>.vlist-t{text-align:left;}.css-4okk7a .katex .mtable .col-align-r>.vlist-t{text-align:right;}.css-4okk7a .katex .svg-align{text-align:left;}.css-4okk7a .katex svg{display:block;position:absolute;width:100%;height:inherit;fill:currentColor;stroke:currentColor;fill-rule:nonzero;fill-opacity:1;stroke-width:1;stroke-linecap:butt;stroke-linejoin:miter;stroke-miterlimit:4;stroke-dasharray:none;stroke-dashoffset:0;stroke-opacity:1;}.css-4okk7a .katex svg path{stroke:none;}.css-4okk7a .katex img{border-style:none;min-width:0;min-height:0;max-width:none;max-height:none;}.css-4okk7a .katex .stretchy{width:100%;display:block;position:relative;overflow:hidden;}.css-4okk7a .katex .stretchy::before,.css-4okk7a .katex .stretchy::after{content:'';}.css-4okk7a .katex .hide-tail{width:100%;position:relative;overflow:hidden;}.css-4okk7a .katex .halfarrow-left{position:absolute;left:0;width:50.2%;overflow:hidden;}.css-4okk7a .katex .halfarrow-right{position:absolute;right:0;width:50.2%;overflow:hidden;}.css-4okk7a .katex .brace-left{position:absolute;left:0;width:25.1%;overflow:hidden;}.css-4okk7a .katex .brace-center{position:absolute;left:25%;width:50%;overflow:hidden;}.css-4okk7a .katex .brace-right{position:absolute;right:0;width:25.1%;overflow:hidden;}.css-4okk7a .katex .x-arrow-pad{padding:0 0.5em;}.css-4okk7a .katex .cd-arrow-pad{padding:0 0.55556em 0 0.27778em;}.css-4okk7a .katex .x-arrow,.css-4okk7a .katex .mover,.css-4okk7a .katex .munder{text-align:center;}.css-4okk7a .katex .boxpad{padding:0 0.3em 0 0.3em;}.css-4okk7a .katex .fbox,.css-4okk7a .katex .fcolorbox{box-sizing:border-box;border:0.04em solid;}.css-4okk7a .katex .cancel-pad{padding:0 0.2em 0 0.2em;}.css-4okk7a .katex .cancel-lap{margin-left:-0.2em;margin-right:-0.2em;}.css-4okk7a .katex .sout{border-bottom-style:solid;border-bottom-width:0.08em;}.css-4okk7a .katex .angl{box-sizing:border-box;border-top:0.049em solid;border-right:0.049em solid;margin-right:0.03889em;}.css-4okk7a .katex .anglpad{padding:0 0.03889em 0 0.03889em;}.css-4okk7a .katex .eqn-num::before{counter-increment:katexEqnNo;content:'(' counter(katexEqnNo) ')';}.css-4okk7a .katex .mml-eqn-num::before{counter-increment:mmlEqnNo;content:'(' counter(mmlEqnNo) ')';}.css-4okk7a .katex .mtr-glue{width:50%;}.css-4okk7a .katex .cd-vert-arrow{display:inline-block;position:relative;}.css-4okk7a .katex .cd-label-left{display:inline-block;position:absolute;right:calc(50% + 0.3em);text-align:left;}.css-4okk7a .katex .cd-label-right{display:inline-block;position:absolute;left:calc(50% + 0.3em);text-align:right;}.css-4okk7a .katex-display{display:block;margin:1em 0;text-align:center;}.css-4okk7a .katex-display>.katex{display:block;white-space:nowrap;}.css-4okk7a .katex-display>.katex>.katex-html{display:block;position:relative;}.css-4okk7a .katex-display>.katex>.katex-html>.tag{position:absolute;right:0;}.css-4okk7a .katex-display.leqno>.katex>.katex-html>.tag{left:0;right:auto;}.css-4okk7a .katex-display.fleqn>.katex{text-align:left;padding-left:2em;}.css-4okk7a body{counter-reset:katexEqnNo mmlEqnNo;}.css-4okk7a table{width:-webkit-max-content;width:-moz-max-content;width:max-content;}.css-4okk7a .tableBlock{max-width:100%;margin-bottom:1rem;overflow-y:scroll;}.css-4okk7a .tableBlock thead,.css-4okk7a .tableBlock thead th{border-bottom:1px solid #333!important;}.css-4okk7a .tableBlock th,.css-4okk7a .tableBlock td{padding:10px;text-align:left;}.css-4okk7a .tableBlock th{font-weight:bold!important;}.css-4okk7a .tableBlock caption{caption-side:bottom;color:#555;font-size:12px;font-style:italic;text-align:center;}.css-4okk7a .tableBlock caption>p{margin:0;}.css-4okk7a .tableBlock th>p,.css-4okk7a .tableBlock td>p{margin:0;}.css-4okk7a .tableBlock [data-background-color='aliceblue']{background-color:#f0f8ff;color:#000;}.css-4okk7a .tableBlock [data-background-color='black']{background-color:#000;color:#fff;}.css-4okk7a .tableBlock [data-background-color='chocolate']{background-color:#d2691e;color:#fff;}.css-4okk7a .tableBlock [data-background-color='cornflowerblue']{background-color:#6495ed;color:#fff;}.css-4okk7a .tableBlock [data-background-color='crimson']{background-color:#dc143c;color:#fff;}.css-4okk7a .tableBlock [data-background-color='darkblue']{background-color:#00008b;color:#fff;}.css-4okk7a .tableBlock [data-background-color='darkseagreen']{background-color:#8fbc8f;color:#000;}.css-4okk7a .tableBlock [data-background-color='deepskyblue']{background-color:#00bfff;color:#000;}.css-4okk7a .tableBlock [data-background-color='gainsboro']{background-color:#dcdcdc;color:#000;}.css-4okk7a .tableBlock [data-background-color='grey']{background-color:#808080;color:#fff;}.css-4okk7a .tableBlock [data-background-color='lemonchiffon']{background-color:#fffacd;color:#000;}.css-4okk7a .tableBlock [data-background-color='lightpink']{background-color:#ffb6c1;color:#000;}.css-4okk7a .tableBlock [data-background-color='lightsalmon']{background-color:#ffa07a;color:#000;}.css-4okk7a .tableBlock [data-background-color='lightskyblue']{background-color:#87cefa;color:#000;}.css-4okk7a .tableBlock [data-background-color='mediumblue']{background-color:#0000cd;color:#fff;}.css-4okk7a .tableBlock [data-background-color='omnigrey']{background-color:#f0f0f0;color:#000;}.css-4okk7a .tableBlock [data-background-color='white']{background-color:#fff;color:#000;}.css-4okk7a .tableBlock [data-text-align='center']{text-align:center;}.css-4okk7a .tableBlock [data-text-align='left']{text-align:left;}.css-4okk7a .tableBlock [data-text-align='right']{text-align:right;}.css-4okk7a .tableBlock [data-vertical-align='bottom']{vertical-align:bottom;}.css-4okk7a .tableBlock [data-vertical-align='middle']{vertical-align:middle;}.css-4okk7a .tableBlock [data-vertical-align='top']{vertical-align:top;}.css-4okk7a .tableBlock__font-size--xxsmall{font-size:10px;}.css-4okk7a .tableBlock__font-size--xsmall{font-size:12px;}.css-4okk7a .tableBlock__font-size--small{font-size:14px;}.css-4okk7a .tableBlock__font-size--large{font-size:18px;}.css-4okk7a .tableBlock__border--some tbody tr:not(:last-child){border-bottom:1px solid #e2e5e7;}.css-4okk7a .tableBlock__border--bordered td,.css-4okk7a .tableBlock__border--bordered th{border:1px solid #e2e5e7;}.css-4okk7a .tableBlock__border--borderless tbody+tbody,.css-4okk7a .tableBlock__border--borderless td,.css-4okk7a .tableBlock__border--borderless th,.css-4okk7a .tableBlock__border--borderless tr,.css-4okk7a .tableBlock__border--borderless thead,.css-4okk7a .tableBlock__border--borderless thead th{border:0!important;}.css-4okk7a .tableBlock:not(.tableBlock__table-striped) tbody tr{background-color:unset!important;}.css-4okk7a .tableBlock__table-striped tbody tr:nth-of-type(odd){background-color:#f9fafc!important;}.css-4okk7a .tableBlock__table-compactl th,.css-4okk7a .tableBlock__table-compact td{padding:3px!important;}.css-4okk7a .tableBlock__full-size{width:100%;}.css-4okk7a .textBlock{margin-bottom:16px;}.css-4okk7a .textBlock__text-formatting--finePrint{font-size:12px;}.css-4okk7a .textBlock__text-infoBox{padding:0.75rem 1.25rem;margin-bottom:1rem;border:1px solid transparent;border-radius:0.25rem;}.css-4okk7a .textBlock__text-infoBox p{margin:0;}.css-4okk7a .textBlock__text-infoBox--primary{background-color:#cce5ff;border-color:#b8daff;color:#004085;}.css-4okk7a .textBlock__text-infoBox--secondary{background-color:#e2e3e5;border-color:#d6d8db;color:#383d41;}.css-4okk7a .textBlock__text-infoBox--success{background-color:#d4edda;border-color:#c3e6cb;color:#155724;}.css-4okk7a .textBlock__text-infoBox--danger{background-color:#f8d7da;border-color:#f5c6cb;color:#721c24;}.css-4okk7a .textBlock__text-infoBox--warning{background-color:#fff3cd;border-color:#ffeeba;color:#856404;}.css-4okk7a .textBlock__text-infoBox--info{background-color:#d1ecf1;border-color:#bee5eb;color:#0c5460;}.css-4okk7a .textBlock__text-infoBox--dark{background-color:#d6d8d9;border-color:#c6c8ca;color:#1b1e21;}.css-4okk7a .text-overline{-webkit-text-decoration:overline;text-decoration:overline;}.css-4okk7a.css-4okk7a{color:#2B3148;background-color:transparent;font-family:"Roboto","Helvetica","Arial",sans-serif;font-size:20px;line-height:24px;overflow:visible;padding-top:0px;position:relative;}.css-4okk7a.css-4okk7a:after{content:'';-webkit-transform:scale(0);-moz-transform:scale(0);-ms-transform:scale(0);transform:scale(0);position:absolute;border:2px solid #EA9430;border-radius:2px;inset:-8px;z-index:1;}.css-4okk7a .js-external-link-button.link-like,.css-4okk7a .js-external-link-anchor{color:inherit;border-radius:1px;-webkit-text-decoration:underline;text-decoration:underline;}.css-4okk7a .js-external-link-button.link-like:hover,.css-4okk7a .js-external-link-anchor:hover,.css-4okk7a .js-external-link-button.link-like:active,.css-4okk7a .js-external-link-anchor:active{text-decoration-thickness:2px;text-shadow:1px 0 0;}.css-4okk7a .js-external-link-button.link-like:focus-visible,.css-4okk7a .js-external-link-anchor:focus-visible{outline:transparent 2px dotted;box-shadow:0 0 0 2px #6314E6;}.css-4okk7a p,.css-4okk7a div{margin:0px;display:block;}.css-4okk7a pre{margin:0px;display:block;}.css-4okk7a pre code{display:block;width:-webkit-fit-content;width:-moz-fit-content;width:fit-content;}.css-4okk7a pre:not(:first-child){padding-top:8px;}.css-4okk7a ul,.css-4okk7a ol{display:block margin:0px;padding-left:20px;}.css-4okk7a ul li,.css-4okk7a ol li{padding-top:8px;}.css-4okk7a ul ul,.css-4okk7a ol ul,.css-4okk7a ul ol,.css-4okk7a ol ol{padding-top:0px;}.css-4okk7a ul:not(:first-child),.css-4okk7a ol:not(:first-child){padding-top:4px;} Initial parameters

Initial volume (V₁)

Initial temperature (T₁)

Final parameters

Final volume (V₂)

Final temperature (T₂)

Use pressure and moles (optional)

Define the pressure and see how many moles of atoms or molecules there are in a container.

Pressure (p)

Amount of gas (n)

Chemistry Learner

It's all about chemistry.

• Chemical Bonds
• Chemical Reactions
• Materials Chemistry
• Organic Chemistry
• Periodic Trends
• Periodic Table Groups
• How to Read Periodic Table
• Naming Covalent Compounds Worksheets
• Net Ionic Equation Worksheets
• Types of Chemical Reactions Worksheets
• Word Equations Worksheets
• Valence Electrons Worksheets
• Graphing Periodic Trends Worksheets
• Periodic Trends Ionization Energy Worksheets
• Atomic Structure And Isotopes Worksheets

Charles’ Law

Explanation, equation [1-4], problems and solutions.

Charles’ law is an experimental gas law that describes how gases expand when heated. It gives a formal relationship between temperature and volume. Charles’ law states that the volume occupied by a gas at constant pressure is proportional to its temperature. In other words, if gas is heated by keeping pressure and mass constant, it will expand [1-4] .

French physicist J.A.C. Charles first suggested an empirical relationship between temperature and pressure in 1787.

Increasing the temperature of a gas confined to a particular volume causes individual gas molecules to move faster. As they move faster, the molecules encounter the walls of the container more often and with greater force. As a result, the pressure is increased. However, if the container volume increases, the number of strikes with the walls decreases. The pressure returns to its initial value [1-4] .

Suppose V is the volume of the gas and T is its temperature. According to Charles’ law,

V : Volume of the gas in liters or m 3

T : Temperature of the gas in absolute scale or Kelvin

k : Proportionality constant

The above equation is shown in the graph below.

From the above equation, it is evident that as the temperature increases, the volume also increases. Similarly, if the temperature decreases, the volume decreases.

Charles’ law can be used to compare two states or conditions of a gas. Suppose a gas at temperature T 1 has volume V 1 . It expands or contracts such that its final volume and temperature are V 1 and T 1 , respectively. Then,

V 1 = kT 1 and V 2 = kT 2

Dividing one by the other

V 1 /T 1 = V 2 /T 2

The above equation gives the relationship between the initial and final conditions of the gas.

Here are some examples of Charles’ law in everyday life [5,6] .

• A hot air balloon rises because burning propane heats the air. The air expands, thereby increasing the volume and decreasing the density. The envelope of air inside the balloon is lighter than the air outside, making it easier for the balloon to rise.
• We breathe air and expand the lungs. In winter, due to cold air inside them, the lungs shrink. Hence, running and jogging become challenging to do in winter.
• A pool tube can inflate or shrink. On a cold winter day, the water temperature is near freezing. Hence, the air temperature inside the tube is low, and the tube shrinks. The opposite happens on a hot summer day. The air inside the tube is heated, and its temperature rises. As a result, the tube inflates.
• The dent in a ping pong ball can be repaired by immersing it in warm water. The high temperature of water raises the air temperature inside the ball. The air expands, and its pressure forces repair the dent.
• The volume of air inside a tire is affected by the outside temperature. On a cold day, the low temperature outside reduces the air temperature inside. Hence, the tire is deflated. On a hot day, the reverse happens. The high temperature outside increases the air temperature inside. As a result, the tire inflates.
• A helium balloon behaves similarly to a tire. It crumbles when the balloon is taken out of a house on a cold day. When brought back inside a warm room, the balloon gets back to its original shape.

Problem 1 : What change in volume results if 4 L of oxygen is cooled by 6.0 °C from 120 °C?

T 1 = 120 + 273 = 393 K

T 2 = 114 + 273 = 387 K

From Charles’ law,

Or, V 2 = V 1 x T 2 /T 1

Or, V 2 = 4 L x 387 K/393 K = 3.94 L

ΔV = V 2 – V 1 = 4 L – 3.94 L = 0.06 L

Problem 2 : A balloon is filled to a volume of 3.2 L at a temperature of 25 ˚C. The balloon is then heated to a temperature of 65 ˚C. Assuming the pressure remains constant throughout, find the new volume of the balloon.

V 1 = 3.2 L

T 1 = 25 ˚C = 273 + 25 = 298 K

T 2 = 65 ˚C = 273 + 65 = 338 K

From Charles law,

Or, V 2 = 3.2 L x 338 K/ 298 K = 3.63 L

• Thoughtco.com
• Chem.libretexts.org
• Ch301.cm.utexas.edu
• Cliffsnotes.com
• Chemistrygod.com

Related Articles

Sublimation

Graham’s Law

Crystal Field Theory

Van der Waals Equation

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Trending Topics

© 2024 ( Chemistry Learner )

PhysicsTeacher.in

High School Physics

Numerical Problems based on Charles’ Law with solution

In this post, we will solve numerical problems using Charles ’ Law . The formulas to be used are as follows:

If V1 is the volume of a certain mass of a gas at temperature T1 and V2 is the volume of the same mass of the same gas at temperature T2 at constant pressure, then according to Charles’ law ,

V 1 /T 1 = V 2 /T 2 (mass and pressure constant) …… (1)

Alternate presentation of Charles’ law gives us this formula:

∴ Volume at t °C, V t = Initial volume at 0°C + Increase in volume = V 0 + V 0 x (1/273) x t

V t = V 0 ( 1 + t/273) ………………. (2)

[ Find detailed discussion on Charles’ law here ]

Solving Numerical Problems using Charles’ Law

A sample of gas occupies 1.50 L at 25°C. If the temperature is raised to 60°C, what is the new volume of the gas if the pressure remains constant?

Solution: V1 = 1.50 L

T1 = 273 + 25 = 298 K

T2 = 60 + 273 = 333 K

Since pressure remains constant, therefore, by applying Charles’ law

V1/T1 = V2/T2

or V2 = (V1/ T1) x T2

= [(1.50 L) /(298 K)] x (333 K)

A sample of helium has a volume of 520 mL at 100°C. Calculate the temperature at which the volume will become 260 mL. Assume that pressure is constant.

V1 = 520 mL

V2 = 260 mL

T1 = 100 + 273 = 373 K

Since pressure remains constant, therefore, by applying Charles’ law :

or T2= (T1/ V1) xV2 = (373/520) (260)= 186.5 K

or t = 186.5 – 273 = –86.5°C

At what centigrade temperature will a given volume of a gas at 0°C become double its volume, pressure remaining constant?

Solution: Let the volume of the gas at 0°C be V.

T1 = 273 + 0 = 273 K

Since pressure remains constant, therefore, by applying

Charles’ law, V 1/T1 = V2/T2

T 2 = (T1/ V1) xV2 = (273 x 2V) / V = 546 K

Changing the temperature to centigrade scale,

Temperature = 546 – 273 = 273°C.

On a ship sailing in a pacific ocean where the temperature is 23.4°C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches the Indian ocean, where the temperature is 26.1°C ?

Solution: According to Charles’ law

T1 = 273 + 23.4 = 296.4 K

T2 = 273 + 26.1 = 299.1

V2 = (V1/T1). T2 = (V1 x T2 )/ T1 = 2L x 299.1K / 296.4K = 2.018 L

What is the increase in volume when the temperature of 800 mL of air increases from 27°C to 47°C under constant pressure of 1 bar ?

Solution: Since the amount of gas and the pressure remains constant, Charles’ law is applicable. i.e.

V 1/T1 = V2/T2

V 1 = 800 mL V2 = ?

T1 = 273 + 27 = 300 K

T2 = 273 + 47 = 320 K

V2 = (V1/T1). T2 = (V1 x T2 )/ T1

V 2 = (800x 320) /300 = 853.3 mL

∴ Increase in volume of air = 853.3 – 800 = 53.3 mL

Related Posts:

• Numerical problems based on Gay Lussac’s law or pressure law
• Harder Numerical problems based on SNELL’S LAW - in Light chapter physics
• Numerical Problems based on combined Gas law or Gas equation & ideal gas equation
• Numerical problems based on the efficiency of machines
• Numerical Problems Based on Lever's mechanical advantage
• Numerical problems based on the inclined plane physics - solved

• Chemistry Concept Questions and Answers
• Charles law Questions

Charles Law Questions

Charle’s law is also referred to as the law of volumes. It tells us about the behaviour of gases. Charle’s law states that the volume is directly proportional to the temperature of the gas at constant pressure.

Or, V / T = k

Charles Law Chemistry Questions with Solutions

Q1. Suppose P, V, and T denote the gas’s pressure, volume, and temperature. In that case, the correct representation of Chale’s law is

• V is directly proportional to T (at constant P)
• V inversely proportional to T (at constant P)
• None of the above

Answer: (a), If P, V, and T denote the gas’s pressure, volume, and temperature, then the correct representation of Chale’s law is V is directly proportional to T (at constant P).

Q2. How can we convert a Celsius temperature to Kelvin temperature?

• By adding 37
• By subtracting 37
• By subtracting 273
• By adding 273

Answer: (d), We can convert a Celsius temperature to Kelvin temperature by adding 273 to it.

Q3. Which element should remain constant if Charle’s law is applied to a gas sample?

• Temperature and the number of moles of a gas
• Pressure and the number of moles of a gas
• Volume and the number of moles of a gas
• Pressure only

Answer: (d), If Charle’s law is applied than the pressure of the gas sample should remain constant.

Q4. What is the value of the gas constant R?

• 8.314 J mol -1 K -1
• 0.082 J litre atm
• 0.987 cal mol -1 K -1
• 83 erg mol -1 K -1

Answer: (a), The value of gas constant R is 8.314 J mol -1 K -1 .

Q5. According to Charle’s law, if the temperature of a gas at constant pressure is increased, the volume will also

• Remains the same
• Can’t be determined

Answer: (a), Charle’s law states that volume is directly proportional to the temperature at constant pressure.

Or, V = kT.

Thus, if the system’s temperature increases, its volume will increase.

Q6. What is Charle’s law?

Answer: Charle’s law states that the volume of the gas is directly proportional to the absolute temperature of the gas at constant pressure.

Thus, if the system’s temperature increases, its volume will increase or if the system’s temperature decreases, its volume will decrease.

Q7. Do you encounter any of the applications of Charle’s law in everyday life? If yes, Where?

Answer: Yes, we encounter applications of Charle’s law in everyday life. When we take a volleyball outside on a hot day, the ball expands a bit. As the temperature increases, its volume also increases, leading to the expansion of volleyball. Similarly, the volleyball shrinks on a cold day as the temperature drops; its size also decreases.

Q8. Is Charles Law indirect or direct relation?

Answer: Charle’s law is a direct relation between the temperature and the volume of the gas. When the molecule’s temperature rises, molecules move faster thereby creating more pressure on the gas container. Hence, increasing the volume of the container. If the pressure of the gas container remains constant then the number of the molecules also remains constant.

Q9. Can we use quantities in °C in Charle’s law?

Answer: No, we can not use quantities in °C in Charle’s law. The relationship between volume and temperature will work when the temperature is taken in kelvin while we can use any quantity for the volume of the gas.

Q10. Match the following.

Q11. Calculate the decrease in temperature (in Celsius) when 2.00 L at 21.0 °C is compressed to 1.00 L.

Answer: Given

Initial Volume (V 1 ) = 2 L

Initial Temperature (T 1 ) = 21.0 °C = (21 + 273) K = 294 K

Final Volume (V 2 ) = 1 L

To Find: Final Temperature (T 2 ) = ?

We can calculate the final temperature of the gas using Charle’s law.

V 1 / T 1 = V 2 / T 2

2 / 294 = 1 / T 2

T 2 = 294 / 2

T 2 = 147 K

T 2 = (147 – 273) = – 126 °C

Hence the final temperature of the gas at volume 1 L is equivalent to – 126 °C.

Q12. A gas occupies a volume of 600.0 mL at a temperature of 20.0 °C. What will be its volume at 60.0 °C?

Initial Volume (V 1 ) = 600.0 mL

Initial Temperature (T 1 ) = 20.0 °C = (20 + 273) K = 293 K

Final Temperature (T 2 ) = 60.0 °C = (60 + 273) K = 333 K

To Find: Final Volume (V 2 ) = ?

We can calculate the final volume of the gas using Charle’s law.

600 / 293 = V 2 / 333

V 2 = (600 X 333) / 293

V 2 = 199800 / 293

V 2 = 681.91 ≈ 682 mL

Hence the final volume of the gas at 60.0 °C is equivalent to 681.91 ≈ 682 mL.

Q13. A gas occupies a volume of 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?

Initial Volume (V 1 ) = 900.0 mL

Initial Temperature (T 1 ) = 27.0 °C = (27 + 273) K = 300 K

Final Temperature (T 2 ) = 132.0 °C = (132 + 273) K = 405 K

900 / 300 = V 2 / 405

V 2 = (900 X 405) / 300

V 2 = 364500 / 300

V 2 = 1215 mL

Hence the final volume of the gas at 132.0 °C is equivalent to 1215 mL or 1.215 L.

Q14. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?

Initial Volume (V 1 ) = 60.0 mL

Initial Temperature (T 1 ) = 33.0 °C = ( 33 + 273) K = 306 K

Final Temperature (T 2 ) = 5.0 °C = ( 5 + 273) K = 278 K

60 / 306 = V 2 / 278

V 2 = ( 60 X 278) / 306

V 2 = 16680 / 306

V 2 = 54.50 mL

Change in the volume = 60.0 – 54.5 = 5.5 mL.

Hence the change in the volume of the gas at 5.0 °C is equivalent to 5.5 mL.

Q15. A gas occupies a volume of 300.0 mL at a temperature of 17.0 °C. What is the volume at 10.0 °C?

Initial Volume (V 1 ) = 300.0 mL

Initial Temperature (T 1 ) = 17.0 °C = (17 + 273) K = 290 K

Final Temperature (T 2 ) = 10.0 °C = ( 10 + 273) K = 283 K

300 / 290 = V 2 / 283

V 2 = (300 X 283) / 290

V 2 = 84900 / 290

V 2 = 292.75 mL

Hence the final volume of the gas at 10.0 °C is equivalent to 292.75 mL.

Practise Questions on Charles Law

Q1. Differentiate between Boyle’s law and Charle’s law.

Q2. A gas occupies a volume of 500.0 mL at a temperature of 10.0 °C. What will be its volume at 50.0 °C?

Q3. A gas occupies a volume of 100.0 mL at a temperature of 27.0 °C. What is the volume at 10.0 °C?

Q4. What change in volume results if 10.0 mL of gas is cooled from 33.0 °C to 15.0 °C?

Q5. A gas occupies a volume of 1 L at a temperature of 17.0 °C. What is the volume at 10.0 °C?

Click the PDF to check the answers for Practice Questions. Download PDF

Recommended Videos

Ideal gas equation definitions, derivation.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

• Illustrations
• Problem Sets
• Calculators

You are here

Charles' law example problems.

• Charles' Law
• Boyle's Law
• Charles' Law Concepts
• Charles' Law Calculations
• Boyle's Law Concepts
• Boyle's Law Calculations

Search form

Latest illustrations.

• Ocean Gyres and Geostrophic Flow
• Langmuir Circulation
• Test sensitivity - specificity calculator
• How earthquakes show us the inside of the Earth
• Surface currents, the Ekman spiral, and Ekman transport

© Andrew Staroscik 2011-2022

• Anatomy & Physiology
• Astrophysics
• Earth Science
• Environmental Science
• Organic Chemistry
• Precalculus
• Trigonometry
• English Grammar
• U.S. History
• World History

... and beyond

• Socratic Meta
• Featured Answers

• Charles' Law

Key Questions

Charles' law can be used to solve a gas law problem involving volume and temperature.

Here is a video discussing an example of solving a gas problem using Charles' law.

Video from: Noel Pauller

Since pressure is kept constant, the only variable that is manipulated is temperature. This means that we can use Charles's law in order to compare volume and temperature. Since volume and temperature are on opposite sides of the ideal gas law, they are directly proportional to one another. As one variable increases, the other will increase as well.

Charles's law is written as follows:

To use this law, we must first convert the temperatures to Kelvin.

Use these temperatures and the initial volume to solve for the final volume.

Example Question #9 : Gases And Gas Laws

Which law is the following formula?

Charles's law

Ideal gas law

Boyle's law

Gay-Lussac's law

Combined gas law

Charles's law defines the direct relationship between temperature and volume. When the parameters of a system change, Charles's law helps us anticipate the effect the changes have on volume and temperature.

Example Question #2 : Using Charles's Law

Charles's law of gases indicates that, at a constant pressure, the volume of a gas is proportional to the temperature. This is calculated by the following equation:

Our first step to solving this equation will be to convert the given temperatures to Kelvin.

Using these temperatures and the initial volume, we can solve for the final volume of the gas.

Our first step to solving this equation will be to convert the given temperature to Kelvin.

Using this temperature and the given volumes, we can solve for the final temperature of the gas.

Example Question #4 : Using Charles's Law

The graph depicted below represents which of the gas laws?

Newton's third law

The graph shows that there is a directly proportional relationship between the volume of a gas and temperature in Kelvin when kept at a constant pressure. This is known as Charles’s law and can be represented mathematically as follows:

Gay-Lussac's law shows the relationship between pressure and temperature. Boyle's law shows the relationship between pressure and volume. Newton's third law is not related to gas principles and states that for every force on an object, there is an equal and opposite force of the object on the source of force.

We expect the volume to increase since volume and temperature are directly proportional. We know that if we heat something the material will expand so we shouldn't get a value that is smaller than our initial volume. Charles Law says that

where the stuff on the left is the initial volume and temperature and the stuff on the right is the final volume and temperature. First off, we MUST convert the temperatures to Kelvin to use Charles Law. This gives

Solving for the final volume,

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

• Terms & conditions
• Privacy policy

© 2023 Sorting Hat Technologies Pvt Ltd

Charles’ Law Problem

Charles' Law is a gas law that defines how gases expand when heated in an experimental setting. In this article we will solve some Charles' Law problems.

Charles’ law is applicable only at low pressures and high temperatures. Only perfect gases are subject to Charles’ law. At high pressures, the relationship between quantity and temperature isn’t linear. In Charles’ law, the pressure is maintained at a constant level. If the pressure remains constant, the volume of the gas is proportional to the temperature. 

Charles’ Law

According to Charles’ law, the volume of a particular gas is proportional to the temperature when the pressure is kept constant. 

V T = k is the Charles’ law equation.

As a result, V = kT. The law is frequently expressed as V=kT when comparing an equivalent substance under two different sets of circumstances.

V 1 V 2 = T 1 T 2

V 1 T 2 = V 2 T 1

As seen in this equation, with the increase in temperature of the gas, its volume also increases. This means Charles’ law may be used to compare volume and temperature. 

At a given temperature (-266.66° C according to Gay-Lussac’s figure), Charles’ law appears to imply that the volume of gas will decrease to zero. Since the gas has no energy at room temperature, the molecules are unable to move. The direct link between temperature and volume is known as Charles’ Law. If the pressure remains constant and the quantity of molecules remains constant while the temperature of the molecules rises, the molecules travel faster, generating greater pressure on the gas container, increasing the volume

Charles’ Law Problems

At 30°C, a gas sample takes up 2.50 L. If the temperature is raised to 70°C and the pressure remains constant, then what will be the new volume of the gas?

Solution: V 1 = 2.50 L

T 1 = 273 + 30 = 303 K

T 2 = 273 + 70 = 343 K

We will be applying Charles’ Law as pressure is constant here:

V 1 /T 1 = V 2 /T 2

or V 2 = (V 1 / T 1 ) x T 2

= [(2.50 L) /(303 K)] x (343 K)

• At 90°C, a helium sample has a volume of 500 mL. Determine the temperature at which the volume of the liquid will become 240 mL. Assume that the pressure stays the same.

Solution: V 1 = 500 mL

V 2 = 240 mL

T 1 = 90 + 273 = 363 K

Since pressure remains constant, therefore, by applying Charles’ law :

or T 2 = (T 1 / V 1 ) xV 2 = (363/500) (240)= 174.2 K

or t = 174.2 – 273 = –98.8°C

• If the pressure constant is given, then find the temperature at which the volume of the gas is tripled at 0 °C?

Solution: Allow V to be the volume of the gas at 0°C.

T 1 = 273 + 0 = 273 K

Because the pressure remains constant, applying

Charles’ law, V 1 /T 1 = V 2 /T 2

T 2 = (T 1 / V 1 ) xV 2 = (273 x 3V) / V = 819 K

Changing the temperature to centigrade scale,

Temperature = 819 – 273 = 546°C.

• A balloon is inflated with 4L air on a ship cruising in the Pacific Ocean with a temperature of 23.4°C. Calculate the volume of the balloon at the location of the Indian Ocean? The temperature of the Indian Ocean to be taken is 26.1°C.

Solution: According to Charles’ law,

T 1 = 273 + 23.4 = 296.4 K

T 2 = 273 + 26.1 = 299.1

V 2 = (V 1 /T 1 ). T 2 = (V 1 x T 2 )/ T 1 = 4L x 299.1K / 296.4K = 4.036 L

• If the temperature of the gas is continuously increasing from 30°C to 50°C, at a constant pressure of 1 bar, then what will be the increase in the volume of gas if its initial volume is 600 ml?

Solution: We will be using Charle’s law here:

V 1 = 600 mL, V 2 = ?

T 2 = 273 + 50 = 323 K

V 2 = (V1/T1). T2 = (V 1 x T 2 )/ T 1

V 2 = (600x 323) /303 = 639.6 mL

∴ Increase in volume of air = 639.6 – 600 = 39.6 mL

• Volume of a given amount of a gas at 47 o C and constant pressure is 225 cm 3 . If the temperature is decreased to 27 o C at constant pressure, then what would be the volume of gas?

Solution: According to Charles’ Law,

V 1 = 225 cm 3

T 1 =273+ 47 = 320 K

T 2 =273+ 27 = 300 K

V 2 = 225×320 300

V 2 = 240 Cm 3

• If 3.00 L of a gas is being compressed to 2.00 L at a given 31.0 °C, what would be the reduction in temperature on the Celsius scale?

(3.00 L) / 304.0 K) = (2.00 L) / (x)

cross multiply to get:

x = 202.6 K

• A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?

(900.0 mL) / (300.0 K) = (x) / (405.0 K)

x = 1215 mL

• Find the change in volume of the given gas, where 40.0 ml of gas is being cooled from 23.0 °C to 2.00 °C?

(40.0 mL) / (296.0 K) = (x) / (275.00 K)

Cross multiply to get:

296x = 11000

x = 37.1 mL 

Change in Volume = 40 – 37.1

The volume decreases by 2.9 mL.

• A gas occupies 2.00 L at standard temperature. What is the volume at 233.0 °C?

In cross-multiplied form, it is this:

V 2 = (V 1 ) [T 2 / T 1 ] <— notice how I grouped the temperatures together

x = (2.00 L) [(506.0 K) / (273.0 K)]

According to Charles’ law, the volume of particular gas is proportional to the temperature when the pressure is kept constant. 

In this article, we have mentioned various Charles’ Law-related questions with solutions to make it easier for students to understand the concept.

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What aspects of Charles' law cannot be changed?

Is charles’ law inverse or direct, what exactly is the point of the charles’ law experiment, what is charles’ law's significance.

In Charles’ law, the pressure is maintained at a constant level. If the pressure remains constant, the volume of the gas is proportional to the temperature, according to Charles’ law.

The direct link between temperature and volume is known as Charles’ Law. If the pressure remains constant and the number of molecules remains constant, while the temperature of the molecules rises, the molecules travel faster, generating greater pressure on the container containing the gas, increasing the volume.

• The goal of this experiment is to investigate how the volume of a gas changes as the temperature changes at constant pressure.

Charles’ Law is a gas law that explains how gases expand when heated in an experimental setting. According to this law, if an amount of gas is held at a constant pressure, the volume and temperature, measured in degrees Kelvin, have a direct relationship.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

6.3 Gas Laws - Boyle's and Charles' Laws

• Last updated
• Save as PDF
• Page ID 171932

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

Learning Objectives

• Learn what is meant by the term gas laws .
• Learn and apply Boyle's law.
• Learn and apply Charles's law.

When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure ( P ) and volume ( V ), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [ n ]), if the temperature ( T ) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related .

There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:

P × V = constant at constant n and T

If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P 1 and V 1 and the new conditions are labeled P 2 and V 2 , we have

P 1 V 1 = constant = P 2 V 2

where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply

P 1 V 1 = P 2 V 2 at constant n and T

This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's law , after the English scientist Robert Boyle, who first announced it in 1662. Figure \(\PageIndex{1}\) shows two representations of how Boyle's law works.

Boyle's law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle's law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won't matter what the unit is, but the unit must be the same on both sides of the equation.

Example \(\PageIndex{1}\)

A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?

First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P 1 and V 1:

P 1 = 2.44 atm and V 1 = 4.01 L

We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for.

P 2 = 1.93 atm and V 2 = ? L

Substituting these values into Boyle's law, we get

(2.44 atm)(4.01 L) = (1.93 atm) V 2

To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit:

\[\frac{(2.44\, atm)(4.01\, L)}{1.93\, atm}=\frac{(1.93\, atm)\, V_{2}}{1.93\, atm}\nonumber \]

Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels:

\[\frac{(2.44\, \cancel{atm})(4.01\, L)}{1.93\, \cancel{atm}}=\frac{(1.93\, \cancel{atm})\, V_{2}}{1.93\, \cancel{atm}}\nonumber \]

What we have left is

\[\frac{(2.44)(4.01\, L)}{1.93}=V_{2}\nonumber \]

Now we simply multiply and divide the numbers together and combine the answer with the \(L\) unit, which is a unit of volume. Doing so, we get \(V_2 = 5.07\, L\)

Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle's law.

Exercise \(\PageIndex{1}\)

If P 1 = 334 torr, V 1 = 37.8 mL, and P 2 = 102 torr, what is V 2 ?

As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.

Example \(\PageIndex{2}\)

A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?

We can still use Boyle's law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters:

\[0.663\, L\times \frac{1000\, ml}{1\, L}=663\, ml\nonumber \]

Now that both volume quantities have the same units, we can substitute into Boyle's law:

\[(722\, torr)(88.8\, ml)=P_{2}(663\, ml)\nonumber \]

\[\frac{(722\, torr)(88.8)\, ml}{(663\, ml)}=P_{2}\nonumber \]

The mL units cancel, and we multiply and divide the numbers to get P 2 = 96.7 torr

The volume is increasing, and the pressure is decreasing, which is as expected for Boyle's law.

Exercise \(\PageIndex{2}\)

If V 1 = 456 mL, P 1 = 308 torr, and P 2 = 1.55 atm, what is V 2 ?

There are other measurable characteristics of a gas. One of them is temperature ( T ). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume ( V ). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related .

A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from Chapter 2 that science uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin , not its temperature in degrees Celsius . If the temperature of a gas is expressed in kelvins, then experiments show that the ratio of volume to temperature is a constant:

\[\frac{V}{T}=constant\nonumber \]

We can modify this equation as we modified Boyle's law: the initial conditions V 1 and T 1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V 2 and T 2 , as long as pressure and the amount of the gas remain constant. Thus, we have another gas law:

\[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\; at\; constant\; P\; and\; n\nonumber \]

This gas law is commonly referred to as Charles's law , after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle's law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale:

K = °C + 273

where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius.

Example \(\PageIndex{3}\)

A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas.

First, we assign the given values to their variables. The initial volume is V 1 , so V 1 = 34.8 mL, and the initial temperature is T 1 , so T 1 = 315 K. The temperature is increased to 559 K, so the final temperature T 2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles's law yields

\[\frac{34.8\, ml}{315\, K}=\frac{V_{2}}{559\, K}\nonumber \]

We solve for V 2 by algebraically isolating the V 2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side:

\[\frac{(559\cancel{K})(34.8\, ml)}{315\, \cancel{K}}=\frac{V_{2}(\cancel{559\, K})}{\cancel{559\, K}}\nonumber \]

The expression simplifies to

\[\frac{(559)(34.8\, ml)}{315}=V_{2}\nonumber \]

By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is

V 2 = 61.8 mL

Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does.

Exercise \(\PageIndex{3}\)

If V 1 = 3.77 L and T 1 = 255 K, what is V 2 if T 2 = 123 K?

It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles's law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles's law. That is, rather than write it as

\[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\nonumber \]

write the equation as

\[\frac{T_{1}}{V_{1}}=\frac{T_{2}}{V_{2}}\nonumber \]

It is still an equality and a correct form of Charles's law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler.

Example \(\PageIndex{4}\)

A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must the temperature of the gas be for its volume to be 25.0 L?

Here, we are looking for a final temperature, so we will use the reciprocal form of Charles's law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins:

−67°C + 273 = 206 K

In using the gas law, we must use T 1 = 206 K as the temperature. Substituting into the reciprocal form of Charles's law, we get

\[\frac{206\, K}{34.8\, L}=\frac{T_{2}}{25.0\, L}\nonumber \]

Bringing the 25.0 L quantity over to the other side of the equation, we get

\[\frac{(25.0\cancel{L})(206\, K)}{34.8\cancel{L}}=T_{2}\nonumber \]

The L units cancel, so our final answer is T 2 = 148 K

This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example.

Exercise \(\PageIndex{4}\)

If V 1 = 623 mL, T 1 = 255°C, and V 2 = 277 mL, what is T 2 ?

235 K, or −38°C

• The behavior of gases can be modeled with gas laws.
• Boyle's law relates a gas's pressure and volume at constant temperature and amount.
• Charles's law relates a gas's volume and temperature at constant pressure and amount.
• In gas laws, temperatures must always be expressed in kelvins.

(40.0 mmHg) (12.3 liters) = (60.0 mmHg) (x) x = 8.20 L Note three significant figures.

(1.00 atm) ( 3.60 liters) = (2.50 atm) (x) x = 1.44 L

(400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot) x = 133 atm

(1.56 L) (1.00 atm) = (3.00 atm) (x) 0.520 L

(11.2 liters) (0.860 atm) = (x) (15.0 L) x = 0.642 atm

(745.0 mmHg) (500.0 mL) = (760.0 mmHg) (x) x = 490.1 mL

(740.0 mmHg) (350.0 mL) = (760.0 mmHg) (x)

(63.0 atm) (338 L) = (1.00 atm) (x)

(166.0 kPa) (273.15 mL) = (101.325 kPa) (x)

(18.0 mmHg) (77.0 L) = (760.0 mmHg) (x)

Volume will decrease.

It will double in size.

(0.755 atm) (4.31 liters) = (1.25 atm) (x)

(8.00 atm) (600.0 mL) = (2.00 atm) (x)

(800.0 torr) (400.0 mL) = (1000.0 torr) (x)

100 °C is to 101.3 kPa as 88 °C is to x x = 89.144 kPa

P 1 V 1 = P 2 V 2 (101.3) (2.0) = (88.144) (x) x = 2.27 L The balloon will not burst.

(1.00 atm) (2.00 L) = (x) (5.00 L) x = 0.400 atm (1.50 atm) (3.00 L) = (y) (5.00 L) y = 0.900 atm

0.400 atm + 0.900 atm = 1.30 atm

(1.00) (2.00) = n 1 RT in the first bulb moles gas = n 1 = 2.00/RT (1.50) (3.00) = n 2 RT in the second bulb moles gas = n 2 = 4.50/RT

total volume = 2.00 + 3.00 = 5.00 (P 3 ) (5.00) = (n 1 + n 2 )RT (P 3 ) (5.00) = (2.00/RT + 4.50/RT)RT (P 3 ) (5.00) = 6.50 P 3 = 6.50 / 5.00 = 1.30 atm