A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K)

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Solution

Q.
A real gas within a closed chamber at 27°C undergoes the cyclic process as shown in figure. The gas obeys $P V^{3} = R T$ equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K) [2024]

1 225 J

2 205 J

3 20 J

4 - 20 J

Ans.
(2)

For A to B:

Given $P V^{3} = R T \Rightarrow P = \frac{R T}{V^{3}}$

Work done $W_{A B} = \int P d V$

$W_{A B} = \int \frac{R T}{V^{3}} d V = R T {[\frac{V^{- 2}}{- 2}]}_{V_{A} = 2}^{V_{B} = 4}$

$= - \frac{R T}{2} [\frac{1}{V^{2}}] = - \frac{R T}{2} [\frac{1}{16} - \frac{1}{4}]$

$= - \frac{R T}{2} (\frac{- 3}{16}) = \frac{3 R T}{32}$

$\Rightarrow W_{A B} = \frac{3}{32} \times 8 \times 300 = 225 J$

$W_{B C}$ (Isobaric process)

$W_{B C} = P Δ V = 10 (2 - 4) = - 20 J$

$W_{A C}$ (Isochoric process) = 0

$W_{net} = W_{A B} + W_{B C} + W_{C A} = 225 - 20 + 0 = 205 J$

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