A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through 1K, then the percentage increase in its pressure will be

Options

A. `0.4%`
B. `0.2%`
C. `0.1%`
D. `0.8%`

Answer

To find the percentage increase in pressure when the temperature of a gas in a closed vessel is increased by 1 K, we can use the ideal gas law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume (constant for a closed vessel)
- \( n \) = number of moles of gas (constant)
- \( R \) = ideal gas constant
- \( T \) = absolute temperature in Kelvin

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the initial temperature \( T_i = 250 \, K \).
- The initial pressure is \( P_i \).

2. **Final Conditions**:
- After heating, the new temperature \( T_f = T_i + 1 = 250 + 1 = 251 \, K \).
- The final pressure is \( P_f \).

3. **Using the Ideal Gas Law**:
- For the initial state:
\[ P_i V = n R T_i \]
- For the final state:
\[ P_f V = n R T_f \]

4. **Dividing the Equations**:
- Since the volume \( V \) and the number of moles \( n \) are constant, we can divide the two equations:
\[ \frac{P_f}{P_i} = \frac{T_f}{T_i} \]

5. **Substituting the Temperatures**:
- Substitute \( T_f \) and \( T_i \):
\[ \frac{P_f}{P_i} = \frac{251}{250} \]

6. **Finding the Change in Pressure**:
- Rearranging gives:
\[ P_f = P_i \cdot \frac{251}{250} \]

7. **Calculating the Increase in Pressure**:
- The increase in pressure \( \Delta P \) is:
\[ \Delta P = P_f - P_i = P_i \cdot \frac{251}{250} - P_i = P_i \left( \frac{251 - 250}{250} \right) = P_i \cdot \frac{1}{250} \]

8. **Calculating the Percentage Increase**:
- The percentage increase in pressure is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta P}{P_i} \right) \times 100 = \left( \frac{P_i \cdot \frac{1}{250}}{P_i} \right) \times 100 = \frac{1}{250} \times 100 = 0.4\% \]

### Final Answer:
The percentage increase in pressure when the gas is heated through 1 K is **0.4%**.