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(A) For an isobaric process,the work done is given by $\Delta W = P \Delta V = P_0(2V_0 - V_0) = P_0 V_0$. Since the pressure and volume change are id

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$\Delta Q_1 - \Delta Q_2 = \Delta U_1 - \Delta U_2$

Vedclass · Answer

(A) For an isobaric process,the work done is given by $\Delta W = P \Delta V = P_0(2V_0 - V_0) = P_0 V_0$. Since the pressure and volume change are identical for both gases,$\Delta W_1 = \Delta W_2 = P_0 V_0$.From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Thus,$\Delta Q_1 - \Delta U_1 = \Delta W_1$ and $\Delta Q_2 - \Delta U_2 = \Delta W_2$. Since $\Delta W_1 = \Delta W_2$,we have $\Delta Q_1 - \Delta U_1 = \Delta Q_2 - \Delta U_2$,which implies $\Delta Q_1 - \Delta Q_2 = \Delta U_1 - \Delta U_2$. Therefore,option $A$ is correct.For a monoatomic gas,$C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$. For a diatomic gas,$C_V = \frac{5}{2}R$ and $C_P = \frac{7}{2}R$. Using the ideal gas law $PV = nRT$,we have $P_0 V_0 = nR T_i$ and $P_0(2V_0) = nR T_f$,so $nR \Delta T = P_0 V_0$.Change in internal energy: $\Delta U_1 = n C_{V1} \Delta T = \frac{3}{2} P_0 V_0$ and $\Delta U_2 = n C_{V2} \Delta T = \frac{5}{2} P_0 V_0$. Clearly,$\Delta U_2 > \Delta U_1$,so option $C$ is correct.Heat given: $\Delta Q_1 = n C_{P1} \Delta T = \frac{5}{2} P_0 V_0$ and $\Delta Q_2 = n C_{P2} \Delta T = \frac{7}{2} P_0 V_0$. Since $\Delta Q_2 > \Delta Q_1$,and $\Delta Q = \Delta U + \Delta W$,it follows that $\Delta U_2 + \Delta W_2 > \Delta U_1 + \Delta W_1$. Thus,option $B$ is incorrect. Since $A$ and $C$ are correct but $B$ is not,the correct choice is not 'All of these'.

Column $I$	Column $II$
$(p)$ isobaric	$(x)$ $\frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$
$(q)$ isothermal	$(y)$ $PV$
$(r)$ adiabatic	$(z)$ $PV \ln 2$

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