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Two moles of an ideal monoatomic gas occupies a volume $V$ at $27^o C$. The gas expands adiabatically to a volume $2\ V$. Calculate $(a)$ the final temperature of the gas and $(b)$ change in its internal energy.
$(a)$ $195 $ $K$ $(b)$ $-2.7$ $kJ$
$(a)$ $189$ $K$ $(b)$ $-2.7$ $kJ$
$(a)$ $195$ $K$ $(b)$ $2.7$ $kJ$
$(a)$ $189$ $ K$ $(b)$ $2.7$ $kJ$
Solution
In an adiabatic process
$T{V^{\gamma – 1}}=constant$ or ${T_1}{V_1}^{\gamma – 1} = {T_2}{V_2}^{\gamma – 1}$
For monoatomic gas $\gamma = \frac{5}{3}$
$\left( {300} \right){V^{2/3}} = {T_2}{\left( {2V} \right)^{2/3}} \Rightarrow {T_2} = \frac{{300}}{{{{\left( 2 \right)}^{2/3}}}}$
${T_2} = 189\,K\,\left( {final\,temperature} \right)$
Change in internal energy $\Delta U = n\frac{f}{2}R\,\Delta T$
$ = 2\left( {\frac{3}{2}} \right)\left( {\frac{{25}}{3}} \right)\left( { – 111} \right) = – 2.7\,kJ$
