Solved thermodynamics exercise

Work in a polytropic transformation $P V^{k} = const$

Exercise 22 · Lesson 5 — Application to the Ideal Gas

polytropic transformation
pressure work
Laplace's law

Statement

Consider a quasi-static transformation of an ideal gas such that $P V^{k} = const$ , with $k \in R$ and $k \neq = 1$ .

Calculate $W_{A \to B}$ in terms of $P_{A}, V_{A}, P_{B}, V_{B}$ , and $k$ .
Recover the result for an isobaric transformation ( $k = 0$ ).
What is its value for an isochoric transformation (without using the preceding formula)?
What happens when $k = 1$ ? Recover $W = - n R T ln (V_{B} / V_{A})$ .

Hint

Set

P (V) = C V^{- k}

with

C = P_{A} V_{A}^{k}

, integrate, then use

C V^{1 - k} = P V

Detailed solution

Solution

Question 1. The polytropic relation can be written

P V^{k} = C, P (V) = C V^{- k}, C = P_{A} V_{A}^{k} = P_{B} V_{B}^{k} .

Using the banker's convention, the work received by the gas is

W_{A \to B} = - \int_{V_{A}}^{V_{B}} P (V) d V = - C \int_{V_{A}}^{V_{B}} V^{- k} d V .

For $k \neq = 1$ , an antiderivative of $V^{- k}$ is $V^{1 - k} / (1 - k)$ . Thus,

W_{A \to B} = - \frac{C}{1 - k} (V_{B}^{1 - k} - V_{A}^{1 - k}) = \frac{C V _{A}^{1 - k} - C V _{B}^{1 - k}}{1 - k} .

Now

C V_{A}^{1 - k} = P_{A} V_{A}, C V_{B}^{1 - k} = P_{B} V_{B} .

Finally,

W_{A \to B} = \frac{P _{A} V _{A} - P _{B} V _{B}}{1 - k} = \frac{P _{B} V _{B} - P _{A} V _{A}}{k - 1}

Question 2. For $k = 0$ , the relation $P V^{0} = C$ becomes $P = C$ : the transformation is isobaric. Since $P_{A} = P_{B} = P$ ,

W_{A \to B} = P_{A} V_{A} - P_{B} V_{B} = P (V_{A} - V_{B}) = - P (V_{B} - V_{A}) = - P Δ V .

This recovers the direct expression for isobaric work.

Question 3. For an isochoric transformation,

V_{B} = V_{A}

and

d V = 0

at all times. Consequently,

W_{A \to B} = - \int_{A}^{B} P d V = 0.

A nontrivial isochoric transformation does not correspond to any finite value of $k$ : if $V$ is constant, the relation $P V^{k} = const$ would also require $P$ to be constant. The isochoric transformation can only be regarded as the formal limit $k \to \infty$ .

Question 4. For

k = 1

, the integrand is proportional to

1/ V

: the antiderivative is no longer a power but a logarithm. The integration must therefore be performed separately. The polytropic relation becomes

P V = C .

For a closed ideal gas, $P V = n R T$ ; the temperature is therefore constant and $C = n R T$ . The work received is then

W_{A \to B} = - \int_{V_{A}}^{V_{B}} \frac{n R T}{V} d V = - n R T ln (\frac{V _{B}}{V _{A}}) .

For a compression, $V_{B} / V_{A} < 1$ , so the logarithm is negative and $W_{A \to B} > 0$ , as expected.