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$1 m o l e$ of monatomic ideal gas undergoes a cyclic process shown in the $P V$ diagram. The process $B C$ is isothermal and the temperature at $B$ is $800 K$ . Find the work done by the gas during the process $A B C$ .

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Solution

$B ⟶ C$
Workdone,
$W_{1} = n R T ln \frac{V_{2}}{V_{1}} = 1. R .800 . ln \frac{{2 V}_{0}}{V_{0}} = 800 \times 8.32 ln 2 = 6656 ln 2$
$C ⟶ A$
Workdone,
$W_{2} = P d v$
$= P_{0} (V_{0} - {2 V}_{0}) = - P_{0} V_{0}$
$A ⟶ B$
Isochoric process,
$∴$ $W_{3} = 0$
$∴$ Total work done,
$W = W_{1} + W_{2} + W_{3}$
$= 6656 ln 2 - P_{0} V_{0}$ .

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