Question

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ?

• A. 44565
• B. 44628
• C. 44563
• D. 44569

Answer 1

The Correct Option is D

To solve this problem, we need to apply the principles of effusion as given by Graham's Law. Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Thus, lighter gases effuse faster than heavier gases.

Let's go through the solution step-by-step:

1. We are given that equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both gases can escape.
2. The question asks us to find the fraction of oxygen that escapes in the time it takes for half of the hydrogen to escape.
3. According to Graham's Law, the rate of effusion \( r \) is given by:
4. r \propto \frac{1}{\sqrt{M}}
5. where \( M \) is the molar mass of the gas.
6. For hydrogen (\( H_2 \)), \( M = 2 \, \text{g/mol} \). For oxygen (\( O_2 \)), \( M = 32 \, \text{g/mol} \).
7. The rates of effusion for hydrogen and oxygen can be written as:
8. \frac{r_H}{r_O} = \sqrt{\frac{M_O}{M_H}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4
9. Thus, hydrogen effuses 4 times faster than oxygen.
10. Given that one-half of the hydrogen escapes, it means:
11. \text{Fraction of hydrogen escaped} = \frac{1}{2}
12. Since hydrogen effuses 4 times faster, the time taken for half the hydrogen to escape would mean \( \frac{1}{4} \) of the time it takes for the same fraction of oxygen to escape.
13. Therefore, in the same time interval, one-fourth of the oxygen has escaped:
14. \text{Fraction of oxygen escaped} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}
15. Hence, the correct answer is option \frac{1}{8} or the corresponding option number.

Therefore, the correct answer is 44569.