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 $320K$ and the universal gas constant $=8.0Jmo{l}^{-1}{K}^{-1}$, the decrease in its internal energy, in joule, is ________.

Answer 1

Answer: 3600

In adiabatic process,
$T{V}^{\gamma -1}=$ constant
$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
For monoatomic gas, $\gamma =\frac{5}{3}$
$\therefore 320\times {\left(V_1\right)}^{\frac{5}{3}-1}=T_2{\left(64V_1\right)}^{\frac{5}{3}-1}$
$\therefore T_2=\frac{320}{(64{)}^{\frac{2}{3}}}=\frac{320}{16}=20K$
$\therefore \Delta U=n{C}_V\Delta T$
$=1\times \frac{3}{2}\times 8\times (320-20)$
$......(\because C_v=\frac{3}{2}R$ for monoatomic gas $)$
$\therefore \Delta U=3600J$