The pressure in vessel that contained pure oxygen dropped from 2000 torr to 1500 torr in 40 min as the oxygen leaked through a small hole into a vacuum. When the same vessel was filled with another gas, the pressure dropped from 2000 torr to 1500 torr in 80 min. What is the molecular weight of the second gas ?

Answer

To solve the problem step by step, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For oxygen (O2):
- Initial pressure (P1) = 2000 torr
- Final pressure (P2) = 1500 torr
- Time taken (t1) = 40 min
- For the unknown gas (let's call it gas A):
- Initial pressure (P1) = 2000 torr
- Final pressure (P2) = 1500 torr
- Time taken (t2) = 80 min

2. **Calculate the Change in Pressure:**
- Change in pressure for both gases:
\[ \Delta P = P1 - P2 = 2000 \text{ torr} - 1500 \text{ torr} = 500 \text{ torr} \]

3. **Calculate the Rate of Effusion:**
- The rate of effusion (R) can be defined as:
\[ R = \frac{\Delta P}{t} \]
- For oxygen (R1):
\[ R_1 = \frac{500 \text{ torr}}{40 \text{ min}} = 12.5 \text{ torr/min} \]
- For gas A (R2):
\[ R_2 = \frac{500 \text{ torr}}{80 \text{ min}} = 6.25 \text{ torr/min} \]

4. **Apply Graham's Law of Effusion:**
- According to Graham's law:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
- Where:
- \(M_1\) = Molar mass of oxygen = 32 g/mol
- \(M_2\) = Molar mass of gas A (unknown)

5. **Substitute the Values:**
- Substitute the rates into Graham's law:
\[ \frac{12.5}{6.25} = \sqrt{\frac{M_2}{32}} \]
- Simplifying the left side:
\[ 2 = \sqrt{\frac{M_2}{32}} \]

6. **Square Both Sides:**
- Squaring both sides gives:
\[ 4 = \frac{M_2}{32} \]

7. **Solve for Molar Mass of Gas A:**
- Rearranging gives:
\[ M_2 = 4 \times 32 = 128 \text{ g/mol} \]

### Final Answer:
The molecular weight of the second gas (gas A) is **128 g/mol**.

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