Chemistry - Thermodynamics - I Question with Solution | TestHub

Using the IUPAC sign convention for thermodynamics, work done is given by $w=-\int pdV$.

For the isothermal reversible process 1 (from volume $V_1$ to $V_2$ at constant temperature $T_1$):
$w_1=-nR{T}_1\ln \left(\frac{V_2}{V_1}\right)$
Given $n=1$ mol and $R=2$ cal K${}^{-1}$ mol${}^{-1}$, we have:
$w_1=-2T_1\ln \left(\frac{V_2}{V_1}\right)$

For the isothermal reversible process 3 (from volume $V_3$ to $V_4$ at constant temperature $T_2$):
$w_3=-nR{T}_2\ln \left(\frac{V_4}{V_3}\right)=-2T_2\ln \left(\frac{V_4}{V_3}\right)$

Adding the work done in these two processes:
$w_1+w_3=-2T_1\ln \left(\frac{V_2}{V_1}\right)-2T_2\ln \left(\frac{V_4}{V_3}\right)$

Let us also verify the other options to be sure.
For option (1): The net work done in a reversible cycle is the area enclosed by the $p-V$ curve, which is non-zero. Thus, $w_1+w_2+w_3+w_4\ne 0$.

For option (3): In adiabatic processes, $q=0$, so from the first law of thermodynamics, $\Delta U=w$.
Therefore, $w_2=\Delta U_2$ and $w_4=\Delta U_4$.
This gives $w_2+w_4=\Delta U_2+\Delta U_4$.
Since internal energy is a state function, for the complete cycle $\Delta U_{cycle}=\Delta U_1+\Delta U_2+\Delta U_3+\Delta U_4=0$.
For isothermal processes 1 and 3, $\Delta U_1=0$ and $\Delta U_3=0$.
Thus, $\Delta U_2+\Delta U_4=0$, which means $w_2+w_4=0$.
However, $\Delta U_2-\Delta U_4=\Delta U_2-(-\Delta U_2)=2\Delta U_2\ne 0$.
Hence, $w_2+w_4\ne \Delta U_2-\Delta U_4$.

For option (4): $w_1+w_2=-2T_1\ln \left(\frac{V_2}{V_1}\right)+\Delta U_2\ne 2T_1\ln \left(\frac{V_2}{V_1}\right)$.

Answer: $w_1+w_3=-2T_1\ln \frac{V_2}{V_1}-2T_2\ln \frac{V_4}{V_3}$