Question

One mole of a monoatomic ideal gas undergoes an adiabatic expansion in which its volume becomes $64$ times its initial value. If the initial temperature of the gas is $320\, K$ and the universal gas constant $=8.0 \,J \,mol ^{-1} K ^{-1}$, the decrease in its internal energy, in joule, is ________.

Answer 1

In adiabatic process,
$ T V^{\gamma-1}=$ constant
$\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$
For monoatomic gas, $\gamma=\frac{5}{3}$
$\therefore 320 \times\left(V_{1}\right)^{\frac{5}{3}-1}=T_{2}\left(64 V_{1}\right)^{\frac{5}{3}-1}$
$\therefore T_{2}=\frac{320}{(64)^{\frac{2}{3}}}=\frac{320}{16}=20 \,K$
$\therefore \Delta U = nC _{ V } \Delta T $
$=1 \times \frac{3}{2} \times 8 \times(320-20)$
$......\left(\because C _{ v }=\frac{3}{2} R\right.$ for monoatomic gas $)$
$\therefore \Delta U =3600 \,J$