Question

1 mol of helium and 3 mol of $N_2$ exert a pressure of 16 atm. Due to a hole in the vessel in which the mixture is placed, the mixture leaks out. Thus, the ratio of rates of helium and nitrogen effusing out initially is:

• A. 1.14
• B. 0.38
• C. 2.65
• D. 0.88

Answer 1

The correct option is D 0.88

Rate of effusion of gas $\left(r\right)$ $\propto$ $P\sqrt{\frac{1}{MM}}$
Where, $P$ is the pressure and $MM$ is the molar mass of gas.

Helium and nitrogen are at different pressures. Then,
$\frac{r_{He}}{r_{N_2}}=\frac{P_{He}}{P_{N_2}}\sqrt{\frac{M_{N_2}}{M_{He}}}$
Total pressure of gases =$\left(P_{N_2+He}\right)=16\text{atm}$

Mole fraction of $He$ $\left({\chi }_{He}\right)$ = $\frac{MolesofHe}{Totalmoles}=\frac{1}{4}$

$P_{He}=P_{Total}\times {\chi }_{He}=16\times \frac{1}{4}=4\text{atm}$
$P_{N_2}=16-4=12\text{atm}$
$\frac{r_{He}}{r_{N_2}}=\frac{4}{12}\sqrt{\frac{28}{4}}=0.88$