Question

The reversible expansion ob an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?

• A. $T_1 = T_2$
• B. $T_3 > T_1$
• C. $W_{\text {isothermal }}>W_{\text {adiabatic }}$
• D. $\Delta U_{\text {isothermal }}>\Delta U_{\text {adiabatic }}$

Answer 1

Answer: (A), (D)

(a) Since, change of state $\left(p_{1}, V_{1}, T_{1}\right)$ to $\left(p_{2}, V_{2}, T_{2}\right)$ is isothermal therefore, $T_{1}=T_{2}^{\prime}$
(b) Since, change of state $\left(p_{1}, V_{1}, T_{1}\right)$ to $\left(p_{3}, V_{3}, T_{3}\right)$ is an adiabatic expansion it brings about cooling of gas, therefore, $T_{3}< T_{ I }$.
(c) Work done is the area under the curve of $p-V$ diagram. As obvious from the given diagram, magnitude of area under the isothermal curve is greater than the same under adiabatic curve, hence $W_{\text {isotherral }} >W_{\text {adiahatic }}$
(d) $\Delta U=n C_{v} \Delta T$
In isothermal process, $\Delta U=0$ as $\Delta T=0$
In adiabatic process. $\Delta U=n C_{1}\left(T_{3}-T_{1}\right)< 0$ as $T_{3}< T_{1}$.
$\Rightarrow \Delta U_{\text {isothermal }}>\Delta U_{\text {adiabatic }}$
NOTE Here only magnitudes of work is being considered otherwise both works have negative sign.