If the pressure of a gas contained in a closed vessel is increased by 0.4 % when heated by 1°C then its initial temperature must be

Options

A. 250 K
B. 250°C
C. 25°C
D. 25 K

Answer

To solve the problem, we will use the relationship between pressure and temperature for a gas in a closed vessel, which is described by Gay-Lussac's law. According to this law, when the volume is constant, the pressure of a gas is directly proportional to its absolute temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Increase in pressure = 0.4% of the initial pressure (P1)
- Increase in temperature = 1°C
- We need to find the initial temperature (T1).

2. **Convert the Percentage Increase in Pressure:**
- The final pressure (P2) can be expressed as:
\[ P2 = P1 + 0.004 \times P1 = 1.004 \times P1 \]

3. **Express the Initial and Final Temperatures:**
- The initial temperature is T1.
- The final temperature (after heating by 1°C) is:
\[ T2 = T1 + 1 \]

4. **Apply Gay-Lussac's Law:**
- According to Gay-Lussac's law:
\[ \frac{P1}{P2} = \frac{T1}{T2} \]
- Substitute the expressions for P2 and T2:
\[ \frac{P1}{1.004 \times P1} = \frac{T1}{T1 + 1} \]

5. **Simplify the Equation:**
- The P1 terms cancel out:
\[ \frac{1}{1.004} = \frac{T1}{T1 + 1} \]

6. **Cross-Multiply to Solve for T1:**
- Cross-multiplying gives:
\[ T1 + 1 = 1.004 \times T1 \]

7. **Rearrange the Equation:**
- Rearranging the equation:
\[ 1 = 1.004 \times T1 - T1 \]
- This simplifies to:
\[ 1 = 0.004 \times T1 \]

8. **Solve for T1:**
- Dividing both sides by 0.004:
\[ T1 = \frac{1}{0.004} = 250 \]

9. **Final Result:**
- The initial temperature (T1) is:
\[ T1 = 250 \text{ °C} \]

### Conclusion:
The initial temperature of the gas in the closed vessel must be **250°C**.