# Gas Laws | iCalculator™

Thermodynamics Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
13.9Gas Laws

In this Physics tutorial, you will learn:

• What is the relationship between pressure and volume of an ideal gas at constant temperature?
• What is the relationship between pressure and temperature of an ideal gas at constant volume?
• What is the relationship between volume and temperature of an ideal gas at constant pressure?
• What is the relationship between volume and amount of an ideal gas at constant temperature and pressure?
• What does the P - V graph represent in each case?

## Introduction

In the previous tutorials, we have pointed out the existence of the three thermodynamic parameters: pressure, volume and temperature. Now we will discuss more in detail what happens when we manipulate any of these variables or keep them constant.

## Relationship between Pressure and Volume at Constant Temperature. The Boyle's Law

Let's consider again the virtual experiment of the previous tutorials with the gas sample enclosed inside the cylinder with a freely moveable piston on top of the gas. We discussed this situation in the previous tutorial when explaining the work done by the gas to move the piston upwards by Δh, which causes a change in gas volume by ΔV. Let's suppose the gas is initially in the state shown above. We have call it the "initial state" of the gas. The gas parameters for this state are denoted as P1, V1 and T1 where P stands for pressure, V for volume and T for temperature.

When we gently press the piston from up, it will lower down causing a compression of the gas. As a result, the gas volume decreases until a new equilibrium is settled. This brings an increase in the gas pressure as the same number of particles collide with a smaller area (the internal area of cylinder walls), given that If we assume the gas as ideal, i.e. no collision between particles does occur; only collisions between particles and cylinder walls are considered, the gas will have some new parameters when the new equilibrium is established. We write them as P2, V2 and T2.

Since in this case we simply push the piston gently down, no change in temperature occurs in gas. Hence, we can write just T for both states instead of T1 and T2. Such a process in which the temperature remains constant during the entire event is known as isothermal process ("iso" = constant, while the term "thermal" implies the temperature. Thus, isothermal process means a process with constant temperature).

If we write the ideal gas law for the two above states, we obtain:

P1 × V1 = n × R × T

for the initial state and

P2 × V2 = n × R × T

for the final state of the gas, where n is the number of moles and R is the ideal gas constant (both remain unchanged during the entire process).

When dividing the above equations with each other, we obtain

P1 × V1/P2 × V2 = 1

Rearranging the last formula, we obtain the relationship between pressure and volume at constant temperature. It is known as the Boyle's Law and mathematically, it is written as

P1 × V1 = P2 × V2

### Example 1

A balloon containing 2 L of air is pressed as shown in the figure. As a result, the air inside the balloon is compressed to a volume of 1.25 L. What is the new air pressure inside the balloon (in atm) if this process occurs at standard atmospheric pressure? Assume the air inside the balloon as an ideal gas.

### Solution 1

Since the process occurs without any change in temperature (no heat sources involved), we can apply the Boyle's Law for the two states: initial (1) and final (2) instead of the ideal gas law. Given that the standard atmospheric pressure is 1 atm, we can write the following quantities as clues:

V1 = 2 L
V2 = 1.25 L
T1 = T2 = T
P1 = 1 atm
P2 = ?

From Boyle's Law, we have

P1 × V1 = P2 × V2

Substituting the values, we obtain for the final pressure of air inside the balloon:

1 atm × 2 L = P2 × 1.25 L
P2 = 1 atm × 2 L/1.25 L
= 1.6 atm

The Pressure vs Volume graph for an isothermal process is shown below: This graph if a hyperbola (in math, a hyperbola is the graph of functions of the form y = k / x. It is also known as the inverse variation function, in which when the independent variable x increases by a factor of k, the dependent variable y decreases by the same factor).

In Boyle's Law, the constant k is represented by the expression n × R × T as at constant temperature T we obtain by rearranging the ideal gas formula:

P = n × R × T/V

or

P = constant/V

### Example 2

Determine the pressure and temperature of an ideal gas sample, whose P - V graph is shown below given that 1 mole of an ideal gas at standard atmospheric pressure and normal temperature occupies 22.4 L of space. ### Solution 2

First, we find the final pressure P2 applying the Boyle's Law given that the graph represents an isothermal process. We have P1 = 270 kPa, V1 = 2 L, V2 = 6 L. Thus,

P1 × V1 = P2 × V2
P2 = P1 × V1/V2
= 270 kPa × 2 L/6
= 90 kPa

As for gas temperature, it can be calculated by considering the coordinates of any known point of the graph, as the temperature is constant during the entire process. But first, we have to calculate the number of gas moles in standard atmospheric pressure. We can take either the initial or the final parameters of gas (let's say the initial ones) and the value of standard atmospheric pressure and use again the Boyle's Law to determine the normal volume of gas. Thus, we have

P0 × V0 = P1 × V1
100 kPa × V0 = 270 kPa × 2 L
V0 = 270 kPa × 2 L/100 kPa = 5.4 L

Hence, the number of moles n is

Number of moles = Volume of the gas sample in normal conditions/Volume of 1 mole of ideal gas at normal conditions
= 5.4 L/22.4 L/mol
= 0.241 moles

Thus, taking the values of the initial state P1 = 270 kPa = 270000 Pa and V1 = 2 L = 0.002 m3, we obtain for temperature by rearranging the ideal gas law for this state:

T = P1 × V1/n × R
= 270000 Pa × 0.002 m3/0.241 moles × 8.31 J/mol × K
= 540 J/2 J/K
= 270 K

Remark!

It is scientifically correct to convert all values of pressure, volume and temperature in the standard SI units. However, it is OK even if you use other units for pressure and volume when applying the Boyle's Law if the same unit is used in both states. Only temperature must be definitely converted into Kelvin degrees as gas laws work only when temperature is expressed in Kelvin scale.

## Relationship between Pressure and Temperature at Constant Volume. The Gay-Lussacs's Law

Now, let's consider again the situation described in the previous tutorial, in which the gas is inside a fixed cylinder and an amount of heat energy is supplied to it. We dealt with situation when working out the formula for the change in the internal energy ΔU of gas, or when explaining the concept of molar specific heat at constant volume CV. Since volume is constant in both states, we can write V1 = V2 = V. This process is known as isovolumetric (isohoric), i.e. a process with constant volume.

When we supply some heat energy to the gas, its temperature increases. As a result, the pressure exerted by gas molecules increases as molecules hit the cylinder walls with a greater force. We can use the same approach as in the previous paragraph to find the relationship between pressure and temperature in both cases: initial and final. We have

P1 × V = n × R × T1

for the initial state and

P2 × V = n × R × T2

for the final state of the gas, where n is the number of moles and R is the ideal gas constant (both remain unchanged during the entire process).

Dividing both equations side by side, we have

P1 × V/P2 × V = n × R × T1/n × R × T2

Or

P1/P2 = T1/T2

The above equation is known as the Gay-Lussac's Law.

The corresponding P - V graph for an isovolumetric process is shown below. Obviously, there is no work done in the isovolumetric process, as the gas volume remains constant. This is also evident when analysing the P - V graph. It is a vertical line, so no area is produced by the graph and the horizontal axis.

### Example 3

A pressure cooker contains some food at 27°C in standard atmospheric pressure when it is ready for putting on fire. The food needs 177°C of temperature for an optimal cooking, after which the cooker's valve starts releasing hot steam from the boiling food to prevent the pressure from increasing further. What is the maximum pressure (in atm) the food reaches during the cooking process? ### Solution 3

This is an isovolumetric process as the volume remains constant during the entire event. We have

P1 = Patm = 1 atm
T1 = 27°C = 27 + 273 = 300 K
T2 = 177°C = 177 + 273 = 450 K
V1 = V2 = V
P2 = ?

Applying the Gay-Lussac's Law for the two states 1 and 2 (initial and final), we obtain

P1/P2 = T1/T2
1 atm/P2 = 300 K/450 K
P2 = 1 atm × 450 K/300 K
= 1.5 atm

## Relationship between Volume and Temperature at Constant Pressure. The Charles's Law

If we supply some heat energy to the ideal gas inside the cylinder of the previous paragraphs by allowing the piston to slide freely up or down (here, up), we obtain a process at constant pressure as it is self-regulated automatically in order to balance the atmospheric pressure which acts from above the piston. Such a process - which occurs at constant pressure - is known as isobaric process. We write P1 = P2 = P. If we apply again the ideal gas law for the two states 1 and 2, we obtain

P × V1 = n × R × T1

for the initial state and

P × V2 = n × R × T2

for the final state of the gas, where n is the number of moles and R is the ideal gas constant (both remain unchanged during the entire process).

Dividing both equations side by side, we have

P × V1/P × V2 = n × R × T1/n × R × T2

Or

V1/V2 = T1/T2

The above equation is known as the Charles's Law.

The corresponding P - V graph for an isovolumetric process is shown below. The area under the graph represents the work done by the gas to lift the piston, as explained in the previous tutorial.

### Example 4

An ideal gas contracts at constant pressure from 8 L to 3 L when some heat energy is taken off from it. As a result, the gas temperature becomes -3°C at the end of process. What w the initial temperature of gas in Celsius degree?

### Solution 4

We have the following clues:

P1 = P2 = P
V1 = 8 L = 0.008 m3
V2 = 3 L = 0.003 m3
T2 = - 23°C = - 3 + 273 = 270 K
T1 = ?

Since this is an isobaric process, we apply the Charles's Law to find the initial temperature of gas. We have

V1/V2 = T1/T2
0.008 m3/0.003 m3 = T1/270 K
T1 = 0.008 m3 × 270 K/0.003 m3
= 720 K

When converted into Celsius scale, this value becomes 720 - 273 = 447°C.

## Relationship between the Volume of Ideal Gas and the Number of Molecules. The Avogadro's Law

So far, we have assumed the amount of gas as constant in all processes discussed. Now, let's suppose we add some extra gas in the container by keeping both pressure and temperature constant.

If the gas is ideal, no collisions between particles occur. Therefore, an additional amount of gas (which means an extra number of molecules) simply brings an increase in volume of gas if no change in the other parameters does occur. This phenomenon was first explained by Avogadro, so the relationship between the volume of ideal gas and the number of its molecules is known as the Avogadro's Law. Its mathematical expression is

V1/V2 = n1/n2

where n is the number of moles.

Avogadro's Law is intended for ideal gases but it is valid for real gases if the values of pressure and temperature are low.

From experiment, it has been found that one mole of an ideal gas occupies 22.4 L of volume at normal temperature and standard atmospheric conditions.

### Example 5

A balloon contains 0.125 moles of an ideal gas at normal conditions. If an additional amount of gas is added to the balloon, its volume becomes 11.2 L. How many moles of gas were added in the balloon?

### Solution 5

First, we find the initial volume of gas given the conversion factor

Volume of 1 mole of ideal gas = 22.4 L

We have

1 mole = 22.4 L
0.125 moles = V1

Thus,

V1 = 0.125 moles × 22.4 L/mol
= 2.8 L

Now, using the Avogadro's Law, we obtain for the final number of moles n2:

V1/V2 = n1/n2
2.8 L/11.2 L = 0.125 moles/n2
n2 = 0.125 moles × 11.2 L/2.8 L
= 0.5 moles

Thus, the amount of gas added in the balloon is

= 0.5 moles - 0.125 moles
= 0.375 moles

## Whats next?

Enjoy the "Gas Laws" physics tutorial? People who liked the "Gas Laws" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Thermodynamics Revision Notes: Gas Laws. Print the notes so you can revise the key points covered in the physics tutorial for Gas Laws
3. Thermodynamics Practice Questions: Gas Laws. Test and improve your knowledge of Gas Laws with example questins and answers
4. Check your calculations for Thermodynamics questions with our excellent Thermodynamics calculators which contain full equations and calculations clearly displayed line by line. See the Thermodynamics Calculators by iCalculator™ below.
5. Continuing learning thermodynamics - read our next physics tutorial: Entropy and the Second Law of Thermodynamics

## Physics Calculators

You may also find the following Physics calculators useful.