Question

1 mol $He$ and $3$ mol $N_2$ exertv a pressure of $16$ atm. Due to a hole in the vessel in which mixture in placed, mixture leaks out. What is the composition of mixture effusing out initially ?

• A. 0.22
• B. 0.44
• C. 0.66
• D. 0.88

Answer 1

Answer: (D)

0.88

$\frac{r_1}{r_2}=\frac{p_1}{p_2}=\sqrt{\frac{M_2}{M_1}}$

$P_{He}=x_{He}\cdot p_{\text{total}}$

$=\frac{1}{4}\times 16=4$ atom

$P_{N_2}=x_{N_2}\cdot p_{\text{total}}$

$=\frac{3}{4}\times 16=12$ atom $=\left(p_{\text{total}}^{-}{p}_{He}\right)$

$\frac{r_{He}}{f_{N_2}}=\frac{p_{He}}{p_{N_2}}\sqrt{\frac{M\left(N_2\right)}{M\left(He\right)}}$

$=\frac{4}{12}\sqrt{\frac{28}{4}}=0.88$

Hence, moles of $He$ and $N_2$ effusing out initially are in the ratio $0.88:1$