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.