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 $T_1 ≠ T_2$

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)
$\omega=-n R T \ln \frac{V_{2}}{V_{1}}$
$= V _{2}> V _{1}$
$\omega =- ve$ (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 $=0$
$\omega=(- P ext \times \Delta V )=0$
(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