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 ideal gas law relationship between pressure and temperature. The relationship states that for a fixed amount of gas at constant volume, the ratio of pressure to temperature is constant. This can be expressed as:

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Where:
- \( P_1 \) = initial pressure
- \( T_1 \) = initial temperature
- \( P_2 \) = final pressure
- \( T_2 \) = final temperature

### Step-by-Step Solution:

1. **Define Initial Conditions**:
- Let the initial pressure \( P_1 = 1 \) atm (standard atmospheric pressure).
- The increase in pressure is given as \( 0.4\% \). Therefore, the final pressure \( P_2 \) can be calculated as:
\[ P_2 = P_1 + 0.004 \times P_1 = 1 + 0.004 = 1.004 \text{ atm} \]

2. **Define Final Temperature**:
- The temperature increase is given as \( 1^\circ C \). Thus, the final temperature \( T_2 \) can be expressed as:
\[ T_2 = T_1 + 1 \]

3. **Set Up the Equation**:
- Using the ideal gas law relationship:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Substituting the known values:
\[ \frac{1}{T_1} = \frac{1.004}{T_1 + 1} \]

4. **Cross Multiply**:
- Cross multiplying gives:
\[ 1 \cdot (T_1 + 1) = 1.004 \cdot T_1 \]
This simplifies to:
\[ T_1 + 1 = 1.004 T_1 \]

5. **Rearranging the Equation**:
- Rearranging the equation to isolate \( T_1 \):
\[ 1 = 1.004 T_1 - T_1 \]
\[ 1 = 0.004 T_1 \]

6. **Solve for \( T_1 \)**:
- Dividing both sides by \( 0.004 \):
\[ T_1 = \frac{1}{0.004} = 250 \text{ K} \]

### Conclusion:
The initial temperature \( T_1 \) must be \( 250 \) K.