A gas is kept at `13^(0)C` in a vessel, if the volume of the gas is kept constant and is heated, the pressure will be doubled to its initial pressure at a temperature

Options

A. `572 K`
B. `286 K`
C. `143 K`
D. `73 K`

Explanation

To solve the problem, we will use the relationship between pressure and temperature for a gas at constant volume, which is derived from the ideal gas law. The steps are as follows:

### Step 1: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as \( 13^\circ C \). To convert this to Kelvin, we use the formula:
\[ T_1 = 273 + 13 = 286 \, K \]

### Step 2: Understand the relationship between pressure and temperature
Since the volume is kept constant, we can use the formula that relates pressure and temperature at constant volume:
\[ \frac{T_2}{T_1} = \frac{P_2}{P_1} \]
Where:
- \( T_1 \) is the initial temperature,
- \( T_2 \) is the final temperature,
- \( P_1 \) is the initial pressure,
- \( P_2 \) is the final pressure.

### Step 3: Set up the equation for the final pressure
According to the problem, the final pressure \( P_2 \) is double the initial pressure \( P_1 \):
\[ P_2 = 2P_1 \]

### Step 4: Substitute \( P_2 \) into the equation
Substituting \( P_2 \) into the pressure-temperature relationship gives us:
\[ \frac{T_2}{T_1} = \frac{2P_1}{P_1} = 2 \]

### Step 5: Solve for the final temperature \( T_2 \)
Now we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = 2T_1 \]
Substituting the value of \( T_1 \):
\[ T_2 = 2 \times 286 \, K = 572 \, K \]

### Step 6: Conclusion
The final temperature \( T_2 \) at which the pressure of the gas will double is:
\[ T_2 = 572 \, K \]