Question

Nitrogen gas $N _{2}$ of mass $28\, g$ is kept in a vessel at pressure of $10 \,atm$ and temperature $57^{\circ} C$. Due to leakage of $N _{2}$ gas its pressure falls to $5 \,atm$ and temperature to $27^{\circ} C$. The amount of $N _{2}$ gas leaked out is

• A. $\frac{5}{63} g$
• B. $\frac{63}{5} g$
• C. $\frac{28}{63} g$
• D. $\frac{63}{28} g$

Answer 1

Answer: (B)

Mass $=28 \,g $
$P_{i}=10 \,atm $
$ T_{i}=57^{\circ} C =330 \,K $
$P_{f}=5 \,atm$
$ T_{f}=27^{\circ} C =300\, K$
Volume is kept constant.
$P_{i}=K \times n_{i} T_{i} $ ..... (i)
$P_{f}=K \times n_{f} T_{f}$ ......(ii)
Dividing (i) by (ii)
$\frac{P_{f}}{P_{f}}=\frac{n_{i}}{n_{f}} \frac{T_{i}}{T_{f}} $
$\frac{n_{i}}{n_{f}}=\frac{10}{5} \times \frac{300}{330}$
or $\frac{n_{i}}{n_{f}}=2 \times \frac{10}{11}$
$\frac{n_{i}}{n_{t}}=\frac{20}{11}$
Now $n_{i}=1$ mole of $N _{2}$
$n_{f}=\frac{11}{20}$ moles
or Mass of $N _{2}$ left $=\frac{11}{20} \times 28$
$\therefore $ Quantity released $=28-\frac{11}{20} \times 28 $
$=\frac{9}{20} \times 28=\frac{63}{5} g$