Question

At 27 $^{o}C$ , 500 ml of helium diffuse in 30 minutes. What is the time (in hours) taken for 100 ml of $S{{O}_{2}}$ to diffuse under the same experimental conditions?
A. 240
B. 0.3
C. 0.2
D. 0.4

Answer 1

There is a formula to calculate the time taken by a gas to diffuse and it is as follows.
$\dfrac{{{V}_{1}}{{t}_{2}}}{{{V}_{2}}{{t}_{1}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$
Here ${{V}_{1}}$ = volume of the gas-1
${{V}_{2}}$ = volume of the gas-2
${{M}_{1}}$ = Molecular weight of the gas-1
${{M}_{2}}$ = Molecular weight of the gas-2
${{t}_{1}}$ = time taken by gas-1 to diffuse
${{t}_{2}}$ = time taken by the gas-2 to diffuse

Complete answer:
- In the question it is given to calculate the time taken by the sulphur dioxide gas to diffuse.
- The given data in the question is the volume of the helium gas is 500 ml (${{V}_{1}}$) and the volume of the sulphur dioxide gas is 100 ml ${{V}_{2}}$ .
- The time taken by the helium gas to diffuse is 30 minutes (${{t}_{1}}$ ).
- The molecular weight of the helium gas is 4 (${{M}_{1}}$ ) and the molecular weight of the sulphur dioxide gas is 64 (${{M}_{2}}$ ).
- By substituting all the values in the below formula to get the time (${{t}_{2}}$ ) taken by the sulphur dioxide gas to diffuse.

& \dfrac{{{V}_{1}}{{t}_{2}}}{{{V}_{2}}{{t}_{1}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}} \\\ & \dfrac{500\times {{t}_{2}}}{100\times 30}=\sqrt{\dfrac{64}{4}} \\\ & {{t}_{2}}=\dfrac{120}{5}=0.4hours \\\ \end{aligned}$$ \- Therefore the time taken by the sulphur dioxide gas to diffuse under the same experimental condition is 0.4 hours. **So, the correct option is D.** **Note:** If we know the time taken by a particular volume of a gas then we can calculate the time taken by the other gas to diffuse under the same experimental condition by using the relation between the time, volume and the molecular weight of the gases.