One mole of an ideal gas undergoes a cyclic process, consisting of two isochores and two isobars. Temperature at 1 and 3 equal to $T_{1}$ and $T_{3}$ respectively. The work done by the gas over the cycle, if the point 2 and 4 lie on the same isotherm :-

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TestHub · Accepted Answer

Points 2 and 4 lie on the same isotherm.
$T_{2} = T_{4}$
$P_{2} v_{1} = P_{1} v_{2}$
$P_{1} v_{1} = R T_{1}$
$P_{2} v_{2} = R T_{3}$
Work done is the area enclosed by the cycle.
$W = (P_{2} - P_{1}) (v_{2} - v_{1})$
$W = P_{2} v_{2} + P_{1} v_{1} - P_{2} v_{1} - P_{1} v_{2}$
$P_{2} v_{1} = P_{1} v_{2}$
$W = R T_{3} + R T_{1} - 2 P_{2} v_{1}$
$(P_{2} v_{1})^{2} = (P_{1} v_{1}) (P_{2} v_{2})$
$(P_{1} v_{1}) (P_{2} v_{2}) = R^{2} T_{1} T_{3}$
$P_{2} v_{1} = R T_{1} T_{3}$
$W = R (T_{1} + T_{3} - 2 T_{1} T_{3})$
$W = R (T_{3} - T_{1})^{2}$
Hence option B.

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