charles law problem solving examples

Charles' Law Example Problem

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

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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.
• What Is the Formula for Charles' Law?
• Gay-Lussac's Gas Law Examples
• The Formula for Boyle's Law
• Avogadro's Law Example Problem
• Charles's Law Definition in Chemistry
• Gay-Lussac's Law Definition
• The Combined Gas Law in Chemistry
• Boyle's Law: Worked Chemistry Problems
• Gases - General Properties of Gases
• Boyle's Law Explained With Example Problem
• How to Calculate Density of a Gas
• Ideal Gas Law Example Problem
• The Formula for the Combined Gas Law
• Ideal Gas Law Test Questions
• Gases Study Guide
• Ideal Gas Example Problem: Partial Pressure

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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

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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

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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

Chemistry Steps

General Chemistry

The Charle’s law shows the correlation between the temperature and the volume of an ideal gas . Let’s discuss it by using a pump with a freely moving plunger that is filled with some gas. In the experiment, we are going to heat up the pump and monitor how the pressure and volume are affected. Remember, we assume that the pressure is constant because the plunger can move freely, and it has no weight.

As expected, the gas is expanding, and we can conclude that the volume of a gas increases with temperature:

The volume is directly proportional to the temperature of the gas, so we can write that:

To bring in the equal sign instead of the proportional sign, we introduce a constant:

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

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

Income ∼ number of cars

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income   = price x number of

So, for our experiment, we are not interested too much in the constant, but rather in how it links the temperature and volume of the gas at positions 1 and 2. Because the ratio V/T is constant at any temperature and volume, we can write that:

\[\frac{{{{\rm{V}}_1}}}{{{{\rm{T}}_{\rm{1}}}}}\, = \,{\rm{constant}}\, = \,\frac{{{{\rm{V}}_2}}}{{{{\rm{T}}_{\rm{2}}}}}\]

This is the practical implication of Charle’s law that is used for solving gas problems.

For example ,

What will be the final volume of a 3.50 L sample of nitrogen at 20 °C if it is heated to 200. °C?

Write down what is given and what needs to be determined first:

V 1 = 3.50 L

T 1 = 20 o C

T 2 = 200. o C

Now, before doing anything else, remember to always convert the temperature to Kelvin when solving a gas problem:

So, T 1 = 20 + 273 = 293 K ,  T 2 = 200 + 273 = 473 K  

The question studies the correlation between the volume and the temperature of a gas, so we need to use the Charle’s law.

Write it down and rearrange it to get an expression of V 2 .

\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}\; = \;\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]

\[{{\rm{V}}_{\rm{2}}}\; = \;\frac{{{\rm{473}}\;\cancel{{\rm{K}}}\;{\rm{ \times }}\;{\rm{3}}{\rm{.50}}\;{\rm{L}}}}{{{\rm{293}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.65}}\;{\rm{L}}\]

• The Ideal Gas Law

Notice that in the experiment, we made an assumption that the pressure is constant, and this is what we also do the other gas laws to determine the desired correlation. For example, in the Boyle’s law , we study a constant amount of gas at constant temperature and find that the pressure increase as the volume is decreased.

To combine all the laws together and have the four variables (n, P, V, T) in one place, the Ideal Gas Law equation is obtained:

The R is called the ideal gas constant. Although it has different values and units, you will mostly be using this:

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

The ideal gas law equation is used when you need to find P, V, T, or n , for the system where they do not change .

We will cover the ideal gas law in a separate post with some examples.

How do I know which gas law to use?

You are probably wondering about this question now that every gas law brings a new equation. For this, there is what is called the combined gas law and as long as you remember it, you do not need to remember all the gas laws to solve a problem.

Let’s keep it for another article because there is quite a lot of information in this one.

• Boyle’s Law
• Gay-Lussac’s Law
• Avogadro’s Law
• Celsius or Kelvin
• Ideal-Gas Laws
• Combined Gas Law Equation
• How to Know Which Gas Law Equation to Use
• Molar Mass and Density of Gases
• Graham’s Law of Effusion and Diffusion
• Graham’s Law of Effusion Practice Problems
• Dalton’s Law of Partial Pressures
• Mole Fraction and Partial Pressure of the Gas
• Gases in Chemical Reactions
• Gases-Practice Problems

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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

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High School Chemistry : Using Charles's Law

Study concepts, example questions & explanations for high school chemistry, all high school chemistry resources, example questions, example question #8 : gases and gas laws.

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 #1 : 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 #2 : 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,

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• 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

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5.3: The Simple Gas Laws- Boyle’s Law, Charles’s Law and Avogadro’s Law

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Learning Objectives

• To understand the relationships among pressure, temperature, volume, and the amount of a gas.

Early scientists explored the relationships among the pressure of a gas ( P ) and its temperature ( T ), volume ( V ), and amount ( n ) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method .

The Relationship between Pressure and Volume: Boyle's Law

As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.

The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure \(\PageIndex{1}\). In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure \(\PageIndex{2}\)). A simple plot of \(V\) versus \(P\) gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:

\[PV = \rm constant \label{10.3.1} \]

Dividing both sides by \(P\) gives an equation illustrating the inverse relationship between \(P\) and \(V\):

\[V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{10.3.2} \]

\[V \propto \dfrac{1}{P} \label{10.3.3} \]

where the ∝ symbol is read “is proportional to.” A plot of V versus 1/ P is thus a straight line whose slope is equal to the constant in Equations \(\ref{10.3.1}\) and \(\ref{10.3.3}\). Dividing both sides of Equation \(\ref{10.3.1}\) by V instead of P gives a similar relationship between P and 1/ V . The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. This law in practice is shown in Figure \(\PageIndex{2}\).

At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure

The Relationship between Temperature and Volume: Charles's Law

Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.

The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure \(\PageIndex{3}\)).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure \(\PageIndex{3}\) that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure \(\PageIndex{3}\), plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.

The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).

We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure \(\PageIndex{3}\) is often referred to as Charles’s law and is stated mathematically as

\[V ={\rm const.}\; T \label{10.3.4} \]

\[V \propto T \label{10.3.5} \]

with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.

The Relationship between Amount and Volume: Avogadro's Law

We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure \(\PageIndex{4}\)). This is the historic “Avogadro’s hypothesis.”

A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,

\[V ={\rm const.} \; (n) \label{10.3.6} \]

\[V \propto.n \text{@ constant T and P} \label{10.3.7} \]

This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.

For a sample of gas,

• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)

The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure \(\PageIndex{5}\). Volume increases with increasing temperature or amount, but decreases with increasing pressure.

The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure ( Boyle’s law ), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure ( Charles’s law ), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present ( Avogadro’s law ). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K) , the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature.

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(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