Question

The rate of effusion of hydrogen gas at 880 Torr and 227 °C is 8 torr min inverse. What will be the rate of effusion of oxygen gas at a pressure of 1760 torr and temperature of 227 °C?

Answer 1

Answer : The rate of effusion of oxygen gas is $2\text{ torr }min^{-1}$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$       ..........(1)

where,

$R_1$ = rate of effusion of hydrogen gas = $8\text{ torr }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = ?

$M_1$ = molar mass of hydrogen gas  = 2 g/mole

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{8\text{ torr }min^{-1}}{R_2})=\sqrt{\frac{32g/mole}{2g/mole}}$

$R_2=2\text{ torr }min^{-1}$

Therefore, the rate of effusion of oxygen gas is $2\text{ torr }min^{-1}$