Two gases A and B are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by gas A is 4 times the time taken by gas B for diffusion of equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular masses?

Open in App

Solution

Step-1: Calculate the relation of the rate of diffusion with respect to time
According to Graham’s law of diffusion: It states that the rate of diffusion if different gases are inversely proportional to the square root of their densities at constant temperature and pressure.
Given, $t_{A} = 4 t_{B}$
t_A =time taken by the gas A for the diffusion
t_B=time taken by the gas B for the diffusion
Given, Volume of both the gases is equal
$R a t e o f d i f f u s i o n o f a g a s = \frac{V o l u m e o f t h e g a s}{T i m e t a k e n b y t h e g a s t o d i f f u s e}$
The rate of diffusion of gas A is given by= $r_{A} = \frac{V_{A}}{t_{A}}$
r_A =rate of the diffusion of gas A
V_A =Volume of the gas A
The rate of diffusion of gas B is given by= $r_{B} = \frac{V_{B}}{t_{B}}$
r_B=rate of the diffusion of gas B
V_B=Volume of the gas B
As the volume of two gases is equal, therefore, V_A=V_B=V
$r_{A} = \frac{V}{t_{A}} r_{B} = \frac{V}{t_{B}}$
Given, $t_{A} = 4 t_{B}$
$r_{A} = \frac{V}{4 t_{B}}$
The ratio of the rates of diffusion
$\frac{r_{A}}{r_{B}} = \frac{V}{4 t_{B}} \times \frac{t_{B}}{V} = \frac{1}{4}$ ……(1)
Step-2: Calculate the relation of the rate of diffusion with respect to molecular masses
Relation of the rate of diffusion with a molecular weight
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{d_{B}}{d_{A}}} = \sqrt{\frac{M_{B}}{M_{A}}}$ …….(2)
d_B=density of gas B
M_B=Molecular mass of gas B
d_A=density of gas A
M_A=Molecular mass of gas A
Comparing equations (1) and (2)
$\frac{r_{A}}{r_{B}} = \frac{1}{4} = \sqrt{\frac{M_{B}}{M_{A}}}$
Therefore, ${(\frac{1}{4})}^{2} = \frac{1}{16} = \frac{M_{B}}{M_{A}}$
Step-3: Equate the rate of diffusion obtained with respect to time with the rate of diffusion concerning molecular masses
$\frac{1}{16} = \frac{M_{B}}{M_{A}}$
By rearranging, $\frac{16}{1} = \frac{M_{A}}{M_{B}}$
Therefore, the ratio of the molecular masses of gas A to gas B= 16:1

Suggest Corrections

Similar questions

Q. The gases

A

and

B

are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by gas

A

is four times the time taken by gas

B

for diffusion of equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular weights?

Q. Two gases A and B are kept in two cylinder and allowed to diffuse through a smallhole . If the time taken by gas A is four times the time taken by gas B for difusion of equal volume of gas under similar conditions of temperature and pressure , what will be the ratio of their molecules masses ?

The time taken by for a certain volume of gas to diffuse through a small hole was 2min. Under similar conditions an equal volume of oxygen took 5.65 minute to pass. The molecular mass of the gas is

Q. The time taken for a certain volume of a certain gas to diffuse through a small hole was

2 m i n

5.65 m i n

to pass. The molecular mass of gas (in amu) is:

Q. The time taken for a certain volume of gas to diffuse through a small hole was 2 min. Under similar conditions an equal volume of oxygen took 5.65 minute to pass. The molecular mass of the gas is:

Join BYJU'S Learning Program