Question

$1$ mole of $N{H}_3$ gas at ${27}^{\circ }C$ is expanded in reversible adiabatic condition to make its volume $8$ times $(\gamma =1.33)$. Final temperature and magnitude of work done respectively are:

• A. $150K,900\text{cal}$
• B. $150K,400\text{cal}$
• C. $250K,1000\text{cal}$
• D. $200K,500\text{cal}$

Answer 1

The correct option is A $150K,900\text{cal}$

Temperature and volume are related in reversible adiabatic process as:
$T_2{V}_2^{\gamma -1}=T_1{V}_1^{\gamma -1}$
$\therefore T_2=\frac{T_1{V}_1^{\gamma -1}}{V_2^{\gamma -1}}=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
$=300{\left(\frac{1}{8}\right)}^{\frac{1}{3}}=300\times \frac{1}{2}=150K$
$\frac{C_p}{C_v}=\gamma =\frac{4}{3}$
$\therefore C_v=3R\text{and}{C}_p=4R$
Adiabatic work done
$\Delta w=C_v\Delta T$
$=3\times 2\times (150-300)=-900\text{cal}$