Question

Two moles of an ideal monoatomic gas occupies a volume $V$ at ${27}^{\circ }C$. The gas expands adiabatically to a volume $2V.$ Calculate (a) the final temperature of the gas and (b) change in its internal energy.

• A. $189K$, $2.7kJ$
• B. $195K$, $-2.7kJ$
• C. $189K$, $-2.7kJ$
• D. $195K$, $2.7kJ$

Answer 1

The correct option is C $189K$, $-2.7kJ$

For adiabatic process relation of temperature and volume is,
$$\begin{array}{c}{T}_2{V}_2^{\gamma -1}=T_1{V}_1^{\gamma -1}\\ \Rightarrow T_2(2V{)}^{2/3}=300(V{)}^{2/3}\\ \begin{array}{c}[\gamma =\frac{5}{3}\text{for monoatomic gases}]\\ \Rightarrow T_2=\frac{300}{2^{2/3}}\approx 189K\end{array}\end{array}$$
Also, in adiabatic process,
$\Delta Q=0,\Delta U=-\Delta W$
or $\Delta U=\frac{-nR\left(\Delta T\right)}{\gamma -1}=-2\times \frac{3}{2}\times \frac{25}{3}(300-189)$
$$\begin{array}{cc}\approx -2.7kJ& \\ T_2\approx 189K,\Delta U& \approx -2.7kJ\end{array}$$