An ideal gas is expanded from $\left(p_{1}, V_{1}, T_{1}\right)$ to $\left(p_{2}, V_{2}, T_{2}\right

Question

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

Professor PC Thomas Classes · Answer

$(\mathrm{a}, \mathrm{b}, \mathrm{c}):(\mathrm{a}) w=-P_{	ext {ext }} \Delta V$

In free expansion, external pressure $=0$\
$	herefore \quad w=0$\
From first law of thermodynamics,

$$
\Delta U=q+w
$$

$	herefore \quad \Delta U=q$\
If the expansion is carried out isothermally,\
$q=0 \quad(\because \Delta U=0$ for isothermal process)\
$	herefore \quad$ It is an adiabatic process.\
If the expansion is carried out adiabatically,\
$\Delta U=0 \quad(\because q=0$ for adiabatic process)\
$	herefore \quad$ It is an isothermal process.\

Thus, area under the curve in reversible adiabatic expansion is lesser than in reversible isothermal expansion.\
(c) The work done on the gas is maximum during irreversible compression.\
(d) Internal energy change,\
$\Delta U=n C_{v} \Delta T$\
when gas is expanded reversibly with $T_{1}=T_{2}, \Delta U=0$\
When gas is expanded reversibly under adiabatic condition, $T_{1}>T_{2}$.\
$	herefore \quad \Delta U=-\mathrm{ve}$\