Question

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

• A. $150\text{K},-900\text{cal}$
• B. $150\text{K},-400\text{cal}$
• C. $250\text{K},-1000\text{cal}$
• D. $200\text{K},-800\text{cal}$

Answer 1

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

We know,
For an adiabatic expansion,
$T{V}^{\gamma -1}=$ constant
$\therefore \frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
Let, $T_1$ be $300\text{K}$ here.
Then,
$T_2=300{\left(\frac{V}{8V}\right)}^{\gamma -1}\Rightarrow T_2=300{\left(\frac{V}{8V}\right)}^{1.33-1}\Rightarrow T_2=300{\left(\frac{1}{8}\right)}^{\frac{1}{3}}=150\text{K}$
We know, according to the 1st law of thermodynamics,
$\Delta E=q+w$
Where, $\Delta E=\text{internal energy of the system}$
For an adiabatic expansion, $q=0$
$\therefore \Delta E=w$
Here, $w=\Delta E=n{C}_{v}dT$
$\therefore w=[1\times \frac{R}{\gamma -1}\times (150-300\left)\right]\Rightarrow w=[1\times \frac{2}{1.33-1}\times (-150\left)\right]\text{cal}\Rightarrow w=-(2\times 3\times 150)\text{cal}=-900\text{cal}$