By increasing temperature of a gas by `6^@ C` its pressure increases by 0.4%, at constant volume. Then initial temperature of gas is

Options

A. 1000 K
B. 1500 K
C. 2000 K
D. 750 K

Answer

To solve the problem, we will use the relationship between pressure and temperature for an ideal gas, which states that pressure is directly proportional to temperature when volume is held constant.

### Step-by-step Solution:

1. **Identify the relationship**: For an ideal gas at constant volume, the relationship between pressure (P) and temperature (T) can be expressed as:
\[ P \propto T \]
This means that if the temperature increases, the pressure also increases proportionally.

2. **Express the changes in terms of percentages**: The change in pressure (ΔP) relative to the initial pressure (P) can be expressed as:
\[ \frac{\Delta P}{P} = \frac{\Delta T}{T} \]
where ΔT is the change in temperature.

3. **Convert the percentage increase in pressure**: According to the problem, the pressure increases by 0.4%, which can be expressed as:
\[ \Delta P = 0.004P \]

4. **Substitute the change in temperature**: The change in temperature is given as 6°C. We can substitute this into our equation:
\[ \frac{0.004P}{P} = \frac{6}{T} \]

5. **Simplify the equation**: The P cancels out:
\[ 0.004 = \frac{6}{T} \]

6. **Rearranging to find T**: Rearranging the equation gives:
\[ T = \frac{6}{0.004} \]

7. **Calculate T**: Performing the calculation:
\[ T = 1500 \text{ K} \]

### Conclusion:
The initial temperature of the gas is **1500 K**.