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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?

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.

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 P_{1}, V_{1} and T_{1} 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 P_{2}, V_{2} and T_{2}.

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 T_{1} and T_{2}. 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:

P_{1} × V_{1} = n × R × T

for the initial state and

P_{2} × V_{2} = 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

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

P_{1} × V_{1} = P_{2} × V_{2}

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.

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:

V_{1} = 2 L

V_{2} = 1.25 L

T_{1} = T_{2} = T

P_{1} = 1 atm

P_{2} = ?

From Boyle's Law, we have

P_{1} × V_{1} = P_{2} × V_{2}

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

1 atm × 2 L = P_{2} × 1.25 L

P_{2} = *1 atm × 2 L**/**1.25 L*

= 1.6 atm

P

= 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*

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.

First, we find the final pressure P_{2} applying the Boyle's Law given that the graph represents an isothermal process. We have P_{1} = 270 kPa, V_{1} = 2 L, V_{2} = 6 L. Thus,

P_{1} × V_{1} = P_{2} × V_{2}

P_{2} = *P*_{1} × V_{1}*/**V*_{2}

=*270 kPa × 2 L**/**6*

= 90 kPa

P

=

= 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

P_{0} × V_{0} = P_{1} × V_{1}

100 kPa × V_{0} = 270 kPa × 2 L

V_{0} = *270 kPa × 2 L**/**100 kPa* = 5.4 L

100 kPa × V

V

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

=

= 0.241 moles

Thus, taking the values of the initial state P_{1} = 270 kPa = 270000 Pa and V_{1} = 2 L = 0.002 m^{3}, we obtain for temperature by rearranging the ideal gas law for this state:

T = *P*_{1} × V_{1}*/**n × R*

=*270000 Pa × 0.002 m*^{3}*/**0.241 moles × 8.31 **J**/**mol × K*

=*540 J**/**2 **J**/**K*

= 270 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.

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 C^{V}.

Since volume is constant in both states, we can write V_{1} = V_{2} = 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

P_{1} × V = n × R × T_{1}

for the initial state and

P_{2} × V = n × R × T_{2}

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

Or

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.

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?

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

P_{1} = P_{atm} = 1 atm

T_{1} = 27°C = 27 + 273 = 300 K

T_{2} = 177°C = 177 + 273 = 450 K

V_{1} = V_{2} = V

P_{2} = ?

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

P

= 1.5 atm

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 P_{1} = P_{2} = P. If we apply again the ideal gas law for the two states 1 and 2, we obtain

P × V_{1} = n × R × T_{1}

for the initial state and

P × V_{2} = n × R × T_{2}

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

Or

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.

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?

We have the following clues:

P_{1} = P_{2} = P

V_{1} = 8 L = 0.008 m^{3}

V_{2} = 3 L = 0.003 m^{3}

T_{2} = - 23°C = - 3 + 273 = 270 K

T_{1} = ?

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

T

= 720 K

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

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

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.

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?

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 = V_{1}

0.125 moles = V

Thus,

V_{1} = 0.125 moles × 22.4 *L**/**mol*

= 2.8 L

= 2.8 L

Now, using the Avogadro's Law, we obtain for the final number of moles n_{2}:

n

= 0.5 moles

Thus, the amount of gas added in the balloon is

n_{added} = n_{2} - n_{1}

= 0.5 moles - 0.125 moles

= 0.375 moles

= 0.5 moles - 0.125 moles

= 0.375 moles

A process in which the temperature remains constant during the entire event (T_{1} = T_{2} = T) is known as **isothermal process**. The relationship between pressure and volume at constant temperature is known as the **Boyle's Law** and mathematically it is written as

P_{1} × V_{1} = P_{2} × V_{2}

where the index 1 means "initial state" and 2 means "final state."

The P - V graph of an isothermal process is a hyperbola as pressure and volume vary inversely to each other.

A thermal process in which the volume remains constant (V_{1} = V_{2} = V) is known as **isovolumetric** (isohoric) process. The relationship between pressure and temperature at constant volume is given by the equation

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

The P - V graph for an isovolumetric process is a vertical line.

Any thermal process, which occurs at constant pressure is known as isobaric process. We write P_{1} = P_{2} = P. The relationship between volume and temperature at constant pressure is given by the equation

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

The area under the P - V graph at constant pressure represents the work done by the gas during the expansion (or the work done on the gas during the contraction).

Any additional amount of gas (which means an extra number of molecules) simply brings an increase in volume of gas when no change in the other parameters 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

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.

1) The volume of a gas sample decreases by a factor of 4 when an extra weight is placed above the piston, as shown in the figure.

What is the final pressure of gas in Pa if initially it was under standard atmospheric pressure?

- 25 kPa
- 400 kPa
- 400 000 kPa
- 25 000 kPa

**Correct Answer: B**

2) Five litres of gas expand isobarically to 15 L when temperature becomes 900 K. What was the initial temperature of gas?

- 27°C
- 300°C
- 2700 K
- 2427 K

**Correct Answer: A**

3) A gas sample is heated by 100 K under constant volume in a fixed container. As a result, pressure becomes 1.2 atm from the value of standard atmospheric pressure it initially was. What is the final temperature of gas?

- 500 K
- 120 K
- 100 K
- 600 K

**Correct Answer: D**

We hope you found this Physics tutorial "Gas Laws" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Thermodynamics with our Physics tutorial on Entropy and the Second Law of Thermodynamics.

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