When the temprature of a gas filled in a closed vessel is increased by `1^(@)C`, its pressure increases by 0.4 percent. The initial temperature of gas was

Options

A. 250 K
B. 500 K
C. `250^(@)C`
D. `25^(@)C`

Answer

To solve the problem, we will use the ideal gas law and the relationship between pressure and temperature. Here's the step-by-step solution:

### Step 1: Understand the relationship between pressure and temperature
In a closed vessel, the volume of the gas remains constant. According to the ideal gas law, we have:

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

where:
- \(P_1\) is the initial pressure,
- \(T_1\) is the initial temperature,
- \(P_2\) is the final pressure,
- \(T_2\) is the final temperature.

### Step 2: Define the change in pressure
The problem states that when the temperature is increased by \(1^\circ C\), the pressure increases by \(0.4\%\). Thus, we can express \(P_2\) in terms of \(P_1\):

\[ P_2 = P_1 + 0.004 P_1 = 1.004 P_1 \]

### Step 3: Define the change in temperature
The temperature increases by \(1^\circ C\), which in Kelvin is:

\[ T_2 = T_1 + 1 \]

### Step 4: Substitute the expressions into the ideal gas law
Substituting \(P_2\) and \(T_2\) into the ideal gas law gives us:

\[ \frac{P_1}{T_1} = \frac{1.004 P_1}{T_1 + 1} \]

### Step 5: Cancel \(P_1\) from both sides
Assuming \(P_1 \neq 0\), we can cancel \(P_1\) from both sides:

\[ \frac{1}{T_1} = \frac{1.004}{T_1 + 1} \]

### Step 6: Cross-multiply to solve for \(T_1\)
Cross-multiplying gives:

\[ T_1 + 1 = 1.004 T_1 \]

### Step 7: Rearrange the equation
Rearranging the equation results in:

\[ 1 = 1.004 T_1 - T_1 \]

This simplifies to:

\[ 1 = 0.004 T_1 \]

### Step 8: Solve for \(T_1\)
Now, we can solve for \(T_1\):

\[ T_1 = \frac{1}{0.004} = 250 \text{ K} \]

### Conclusion
The initial temperature of the gas was \(250 \text{ K}\).