One mole of an ideal gas having initial volum | vedclass

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(C) The work done in a cyclic process on a $P-T$ diagram is given by the area enclosed by the cycle. However,we can calculate it by summing the work d

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$RT \ln 2$

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(C) The work done in a cyclic process on a $P-T$ diagram is given by the area enclosed by the cycle. However,we can calculate it by summing the work done in each process: $W_{net} = W_{AB} + W_{BC} + W_{CD} + W_{DA}$.For an ideal gas,$PV = RT$,so $V = \frac{RT}{P}$.Process $AB$: Isobaric $(P = 2P)$,$T$ changes from $T$ to $2T$. Work $W_{AB} = P \Delta V = nR \Delta T = 1 \cdot R(2T - T) = RT$.Process $BC$: Isothermal $(T = 2T)$,$P$ changes from $2P$ to $P$. Work $W_{BC} = nRT \ln(\frac{P_i}{P_f}) = 1 \cdot R(2T) \ln(\frac{2P}{P}) = 2RT \ln 2$.Process $CD$: Isobaric $(P = P)$,$T$ changes from $2T$ to $T$. Work $W_{CD} = nR \Delta T = 1 \cdot R(T - 2T) = -RT$.Process $DA$: Isothermal $(T = T)$,$P$ changes from $P$ to $2P$. Work $W_{DA} = nRT \ln(\frac{P_i}{P_f}) = 1 \cdot R(T) \ln(\frac{P}{2P}) = RT \ln(\frac{1}{2}) = -RT \ln 2$.Total work $W_{net} = RT + 2RT \ln 2 - RT - RT \ln 2 = RT \ln 2$.

One mole of an ideal gas having initial volume $V$,pressure $2P$ and temperature $T$ undergoes a cyclic process $ABCDA$ as shown in the figure. The net work done in the complete cycle is:

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