Question

The reversible expansion of an ideal gas under adiatbatic 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. $w_{isothermal}>w_{adiabatic}$
• D. $\Delta U_{isothermal}>\Delta U_{adiabatic}$

Answer 1

Answer: (A), (C), (D)

$\Delta U_{isothermal}>\Delta U_{adiabatic}$
$w_{isothermal}>w_{adiabatic}$
$T_1=T_2$

As, $(P_1,V_1,T_1)\to (P_2,V_2,T_2)$ is an isothermal process, $\Delta T=0$

$\therefore T_1=T_2$

Work done is equal to are under $P-V$ graph. The area under $P-V$ graph of the isothermal process is greater than that of the adiabatic process.

$\therefore W_{isothermal}>W_{adiabatic}\to \left(i\right)$

Also, $q=0$ for adiabatic process.

$\therefore {\Delta U}_{adiabatic}=W_{adiabatic}$ (for adiabatic process) $\to \left(ii\right)$

& ${\Delta U}_{isothermal}=q_{isothermal}+W_{isothermal}\to \left(iii\right)$

From (i), (ii) and (iii), it can be concluded that

${\Delta U}_{\text{isothermal}}>{\Delta U}_{\text{adiabatic}}$

Hence, the correct options are $\text{A}$, $\text{C}$ and $\text{D}$