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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.

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.

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

=P

=P

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

Volume decreases ⇒ ΔV < 0 ⇒ W

Volume constant ⇒ ΔV = 0 ⇒ W

as we discussed earlier.

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**)

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

m = 12 g

M = 4 *g**/**mol*

t_{1} = 20°C

t_{2} = 100°C

P^{1} = P^{2} = 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

=

= 3 moles

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

∆T = t_{2} - t_{1}

= 100°C - 20°C

= 80°C

= 80 K

= 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 × V_{1} = n × R × T_{1}

and

P × V_{2} = n × R × T_{2}

Subtracting the first equation from the second, we obtain

P × V_{2} - P × V_{1} = n × R × T_{2} - n × R × T_{1}

or

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

= 3 moles × 8.31

= 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.

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