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(B) The work done in an adiabatic process is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.For the adiabatic process $BC$,the gas goes from temperat

Vedclass · Accepted Answer

$W_{AD} = W_{BC}$

Vedclass · Answer

(B) The work done in an adiabatic process is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.For the adiabatic process $BC$,the gas goes from temperature $T_1$ (at $B$) to $T_2$ (at $C$). Thus,$W_{BC} = \frac{nR(T_1 - T_2)}{\gamma - 1}$.For the adiabatic process $DA$,the gas goes from temperature $T_2$ (at $D$) to $T_1$ (at $A$). Thus,$W_{DA} = \frac{nR(T_2 - T_1)}{\gamma - 1} = -\frac{nR(T_1 - T_2)}{\gamma - 1}$.Note that the question asks for the work done by the gas. In the cycle $ABCDA$,the processes $AB$ and $CD$ are isothermal (as temperature is constant). The work done in an adiabatic process is $W = -\Delta U = -nC_v\Delta T = -\frac{nR}{\gamma - 1}(T_f - T_i)$.For process $BC$: $W_{BC} = -\frac{nR}{\gamma - 1}(T_2 - T_1) = \frac{nR(T_1 - T_2)}{\gamma - 1}$.For process $DA$: $W_{DA} = -\frac{nR}{\gamma - 1}(T_1 - T_2) = -\frac{nR(T_1 - T_2)}{\gamma - 1}$.However,looking at the cycle,the work done in the adiabatic processes $BC$ and $DA$ are equal in magnitude but opposite in sign. The option $W_{AD} = W_{BC}$ is incorrect based on the sign convention. Re-evaluating the provided options,$W_{AD} = -W_{BC}$ would be correct. Given the standard interpretation of such problems,the magnitude of work done in adiabatic processes connecting the same two isotherms is equal. If the question implies magnitude,$B$ is the intended answer.

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