Physics Lesson 13.6.4 - Work in a Thermodynamic Process

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Welcome to our Physics lesson on Work in a Thermodynamic Process, this is the fourth lesson of our suite of physics lessons covering the topic of The Kinetic Theory of Gases. Ideal Gases, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Work in a Thermodynamic Process

Let's find an expression for calculation of the work done by a gas during a slow thermal process. Consider a cylinder filled with gas with a freely moveable piston above it as shown in the figure below.

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Since the piston moves freely, the gas pressure remains constant because it is balanced from the atmospheric pressure, i.e. the pressure exerted by the air from above the piston. The work done by the gas to raise the piston by Δy as shown in the figure below, is

Given that

Force = Pressure × Area

we have

W_{by gas}=P_gas × A_piston × ∆y
=P_gas × (A_piston × ∆y)
=P_gas × ∆V

This expression gives the left part of the ideal gas law formula.

The formula indicates that:

Volume increases ⇒ ΔV > 0 ⇒ W_{by gas} > 0
Volume decreases ⇒ ΔV < 0 ⇒ W_{by gas} < 0
Volume constant ⇒ ΔV = 0 ⇒ W_{by gas} = 0

as we discussed earlier.

Example 4

12 g of helium gas is heated from 20 ºC to 100 ºC, at a constant pressure. What is the work done by helium during the heating process? (Take MHe = 4 g/mol)

Solution 4

First, let's write the clues to create a clearer idea about the situation. We have:

m = 12 g
M = 4 g/mol
t₁ = 20°C
t₂ = 100°C
P¹ = P² = P
W_{by gas} = ?

From the first two clues we can work out the number of moles, n. We have:

n = m/M
= 12 g/4 g/mol
= 3 moles

Also, from the next two clues, we obtain the change in temperature:

∆T = t₂ - t₁
= 100°C - 20°C
= 80°C
= 80 K

Applying the ideal gas law for the two given instants (states) 1 and 2, we obtain

P × V₁ = n × R × T₁

and

P × V₂ = n × R × T₂

Subtracting the first equation from the second, we obtain

P × V₂ - P × V₁ = n × R × T₂ - n × R × T₁

P × ∆V = n × R × ∆T

The first part of the last equation gives the work done by the gas. Therefore, the left part will also give the work done by the gas. Hence, we can write:

W_{by gas} = n × R × ∆T
= 3 moles × 8.31 J/mol × K × 80 K
= 1994.4 J

This result means helium does 1994.4 J of work against the environment during thermal expansion.

You have reached the end of Physics lesson 13.6.4 Work in a Thermodynamic Process. There are 6 lessons in this physics tutorial covering The Kinetic Theory of Gases. Ideal Gases, you can access all the lessons from this tutorial below.

More The Kinetic Theory of Gases. Ideal Gases Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.6	The Kinetic Theory of Gases. Ideal Gases
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.6.1	The Meaning of Mole. Avogadro's Number
13.6.2	Ideal Gases
13.6.3	The Ideal Gas Law
13.6.4	Work in a Thermodynamic Process
13.6.5	Finding Work using a P - V Diagram
13.6.6	State Variables and Path-Dependent Variables

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Physics Lesson 13.6.4 - Work in a Thermodynamic Process

Work in a Thermodynamic Process

Example 4

Solution 4

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