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Physics Lesson 13.8.3 - Molar Specific Heat at Constant Pressure

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Welcome to our Physics lesson on Molar Specific Heat at Constant Pressure, this is the third lesson of our suite of physics lessons covering the topic of Molar Specific Heats and Degrees of Freedom, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Molar Specific Heat at Constant Pressure

Now, let's assume the container which holds the ideal gas is not fixed, i.e. its piston can slide freely up and down.

Physics Tutorials: This image provides visual information for the physics tutorial Molar Specific Heats and Degrees of Freedom

If we supply some heat to the gas enough to increase its temperature from T to T + ΔT like in the previous paragraph, the piston will move up as the gas expands due to the increase in temperature but the pressure remains constant. In both cases, the inner pressure of gas balances the effect of atmospheric pressure plus the pressure exerted by the piston's weight).

Similarly as in the process with constant volume, we have for the heat absorbed by the ideal gas at constant pressure

Q = n × Cp × ∆T

where Cp is known as the molar specific heat at constant pressure.

The value of Cp is numerically greater than the corresponding value of CV for the same change in temperature as in the process with constant volume, a part of heat energy supplied goes for doing work for lifting the piston.

Given that the temperature increases without any change in pressure, we obtain for the P - V graph:

Physics Tutorials: This image provides visual information for the physics tutorial Molar Specific Heats and Degrees of Freedom

Let's find the relationship between the two molar specific heats CV and Cp.

From the First Law of Thermodynamics, we know that

∆U = Q - W

Given that at constant pressure the work done by the gas to lift the piston is

W = P × ∆V = n × R × ∆T

we obtain

∆U = Q - n × R × ∆T
= n × Cp × ∆T - n × R × ∆T

Also, given that

∆U = n × Cv × ∆T

we obtain

n × Cv × ∆T = n × Cp × ∆T - n × R × ∆T

Dividing both sides by n × ΔT, we obtain

Cv = Cp - R

Or

Cp = Cv + R

For a monoatomic gas, we have

Cp = 3/2 R + R
= 5/2 R
≈ 20.8 J/mol × K

Therefore, we can write for the heat absorbed by a gas at constant pressure to increase its temperature by ΔT:

∆Q = 5/2 n × R × ∆T

This value is greater than the corresponding value of heat energy during a process at constant volume for the same increase in temperature (Q = 3/2 n R T), as we predicted earlier.

Example 2

What is the heat energy required to increase by 50 K the temperature of 2 moles of an ideal monoatomic gas at constant pressure?

Solution 2

Clues:

ΔT = 50 K
n = 2
ΔQ = ?

Using the equation for heat absorbed by an ideal gas at constant pressure, we can write:

∆Q = 5/2 n × R × ∆T
= 5/2 × 2 moles × 8.31 J/mol × K × 50 K
= 2077.5 J

You have reached the end of Physics lesson 13.8.3 Molar Specific Heat at Constant Pressure. There are 4 lessons in this physics tutorial covering Molar Specific Heats and Degrees of Freedom, you can access all the lessons from this tutorial below.

More Molar Specific Heats and Degrees of Freedom Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
13.8Molar Specific Heats and Degrees of Freedom
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
13.8.1Internal Energy
13.8.2Molar Specific Heat at Constant Volume
13.8.3Molar Specific Heat at Constant Pressure
13.8.4Degrees of Freedom and Molar Specific Heats

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