Question

V ml of H2 gas diffuses through a small hole in a container in time t1. How much time will be required by O2 gas for the diffusion of same volume?

Answer 1

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Answer 2

Answer : The relation between the time of two diffused gas is,  $t_2=4\times t_1$

Solution : Given,

Molar mass of $H_2$ = 2 g/mole

Molar mass of $O_2$ = 32 g/mole

According to the graham's law of diffusion, the rate is inversely proportional to the square root of the mass.

The formula of the rate of diffusion of two gas will become,

$\frac{R_1}{R_2}= \sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of diffusion of $H_2$ gas

$R_2$ = rate of diffusion of $O_2$ gas

$M_1$ = molar mass of $H_2$ gas

$M_2$ = molar mass of $O_2$ gas

Now put all the given values in this formula, we get the relation between the $R_1$ and $R_2$.

$\frac{R_1}{R_2}= \sqrt{\frac{32g/mole}{2g/mole}}=\frac{4}{1}$

In general, the formula of rate of diffusion is,

$\text{ Rate of diffusion}=\frac{\text{ Volume of diffused gas}}{\text{ Time of diffused gas}}$

As per question, the volume of diffused gases are equal.

Now the relation between time and rate of two diffused gas is,

$\frac{R_1}{R_2}= \frac{t_2}{t_1}$

where,

$t_1$ = time of $H_2$ gas

$t_2$ = time of $O_2$ gas

Now put the given values in this formula, we get the relation between the time of two diffused gases.

$\frac{4}{1}= \frac{t_2}{t_1}\\t_2=4\times t_1$

From this, we conclude that the time of diffused gas $O_2$ is four times of the time of diffused gas $H_2$.