Question

An ideal gas is expanded from $\left(p_{1} , V_{1} , T_{1}\right)$ to $\left(\right.p_{2},V_{2},T_{2}\left.\right)$ under different conditions. The incorrect statement among the following is?

• A. The word 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
• B. 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 $T_{1}\neq T_{2}$
• C. If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic
• D. The work done on the gas is maximum when it is compressed irreversibly from $\left(p_{2} , V_{2}\right)$ to $\left(p_{1} , V_{1}\right)$ against constant pressure $p_{1}$

Answer 1

Answer: (B)

Area under curve in reversible isothermal is more. So, more work will be done by gas.

$T_{1}=T_{2}$

$\Delta U=nC_{V}\Delta T=0$

In reversible adiabatic expansion, $T_{2} < T_{1}$

So $\Delta T=-ve$ , $\Delta U=-ve$

In free expansion, $P_{e x t}=0$

So $W=0$

If carried out isothermally $\left(\Delta U = 0\right)$

$q=0$ (adiabatic); from I law

If carried out adiabatically $\left(q = 0\right)$

$\Delta U=0$ (isothermal); From I law



During irreversible compression, maximum work is done on the gas (corresponding to shaded area).