Question

An ideal gas is expanded from ($p_1,V_1,T_1$) to ($p_2,V_2,T_2$) under different conditions. The correct statement(s) among the following is(are)

• A. The work done on the gas is maximum when it is compressed irreversibly from ($p_2,V_2$) to ($p_1,V_1$) against constant pressure $p_1$
• B. If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic
• C. The work done by the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under isothermal conditions
• D. The change in internal energy of the gas is (i) zero, if it is expanded reversibly with $T_1=T_2$, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $T_1\ne T_2$

Answer 1

Answer: (A), (B), (C)

(A) $\omega =-P$ ext $\Delta V$
Work done on gas when compressed
$\omega =$ is positive
$\therefore$ equation becomes
$\omega =P$ ext $\Delta V$
(B) under adiabatic conditions
$\omega =\frac{\left(P_2{V}_2-P_1{V}_1\right)}{(\gamma -1)}$
since equation $V_2>V_1$
$\therefore \omega =+$ ve (work done on system)
for reversibly condition (Isothermal)
$\omega =-nRT\ln \frac{V_2}{V_1}$
$=V_2>V_1$
$\omega =-ve$ (work done on system)
$\omega =-ve$ (for adiabatic reversibly work done by gas)
$\omega =+ve$ (for isothermal reversibly work done by gas)
(C) As free expansion in there (irreserible)
$\therefore P$ ext $=0$
$\omega =(-Pext\times \Delta V)=0$
(Isothermal irreversible process)
In adiabatic $\Delta U=-ve$
$\Delta U=q+\omega$
$-=0+\omega$
$\omega =-$ negative
similarly, $\Delta G=\Delta H-T\Delta s$
gas expand $\therefore \Delta s=+ve$
$\Delta G=\Delta H-Ve$
$=\Delta V+p\Delta V-ve$
$=q+w+p\Delta c-ve$
$=o+w+w-ve$
$=-ve+(-ve)-ve$
$\Delta G=-ve$
$\therefore$ It is spontaneous process
$\therefore$ both take place simultaneously