Question

Nitrogen gas $N_2$ of mass $28g$ is kept in a vessel at pressure of $10atm$ and temperature $57^{\circ }C$. Due to leakage of $N_2$ gas its pressure falls to $5atm$ 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 $=28g$
$P_i=10atm$
$T_i=57^{\circ }C=330K$
$P_f=5atm$
$T_f=27^{\circ }C=300K$
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 $=28-\frac{11}{20}\times 28$
$=\frac{9}{20}\times 28=\frac{63}{5}g$