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Two moles of monoatomic gas is expanded from $(P_0, V_0)$ to $(P_0 , 2V_0)$ under isobaric condition. Let $\Delta Q_1$, be the heat given to the gas, $\Delta W_1$ the work done by the gas and $\Delta U_1$ the change in internal energy. Now the monoatomic gas is replaced by a diatomic gas. Other conditions remaining the same. The corresponding values in this case are $\Delta Q_2 , \Delta W_2 , \Delta U_2$ respectively, then
$\Delta Q_1 - \Delta Q_2 = \Delta U_1 - \Delta U_2$
$\Delta U_2 + \Delta W_2 > \Delta U_1 + \Delta W_1$
$\Delta U_2 > \Delta U_1$
All of these
Solution
since for both cases $P-V$ curve is the same,
$\Delta W_{1}=\Delta W_{2}$ $\Rightarrow \Delta Q_{1}-\Delta U_{1}=\Delta Q_{2}-\Delta U_{2}$
$\Rightarrow \Delta Q_{1}-\Delta Q_{2}=\Delta U_{1}-\Delta U_{2}$
So, option $a$ is correct.
Process is isobaric
Hence $\Delta Q_{1}=n C_{p 1} \Delta T_{1}=\frac{5}{2} n R \Delta T_{1}=\frac{5}{2} P_{0} V_{0}$
Also $\Delta Q_{2}=n C_{P 2} \Delta T_{2}=\frac{7}{2} n R \Delta T_{2}=\frac{7}{2} P_{0} V_{0}$
Hence $\Delta Q_{2}>\Delta Q_{1}$
$\Rightarrow \Delta U_{1}+\Delta W_{1}>\Delta U_{2}+\Delta W_{2}$
So, option $b$ is correct
$\Delta U_{1}=n C_{v 1} \Delta T_{1}=\frac{3}{2} n R \Delta T_{1}=\frac{3}{2} P_{0} V_{0}$
$\Delta U_{2}=n C_{v 2} \Delta T_{2}=\frac{5}{2} n R \Delta T_{2}=\frac{5}{2} P_{0} V_{0}$
Hence $\Delta U_{2}>\Delta U_{1}$
So, option $c$ is also correct
