QJEEM25Phd314: JEE Main 2025 Physics - Class 11 JEEM25Phd314 Solution: A container of fixed volume contains a gas at 27°C. To double…

To solve this problem, we can use the combined gas law in terms of pressure ( $P$ ) and temperature ( $T$ ). For a gas with a fixed volume, the relationship is given as:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

where:

$P_1$ is the initial pressure,

$T_1$ is the initial temperature in Kelvin,

$P_2$ is the final pressure,

$T_2$ is the final temperature in Kelvin.

Given that the initial temperature $T_1 = 27°C$ , we need to convert it to Kelvin:

$T_1 = 27°C + 273.15 = 300.15 \, \text{K}$

The final pressure $P_2$ is twice the initial pressure ( $P_2 = 2P_1$ ). Substituting these values into the gas law equation gives:

$\frac{P_1}{300.15} = \frac{2P_1}{T_2}$

Solving for $T_2$ :

Cancel $P_1$ from both sides:

$\frac{1}{300.15} = \frac{2}{T_2}$

Rearrange to solve for $T_2$ :

$T_2 = 2 \times 300.15 = 600.3 \, \text{K}$

Convert the final temperature back to Celsius:

$T_2 = 600.3 \, \text{K} - 273.15 = 327.15°C$

Therefore, to double the pressure of the gas, the temperature should be raised to approximately 327°C.