QJEEM24Ph2c0d: JEE Main 2024 Physics - Class 11 JEEM24Ph2c0d Solution: The volume of an ideal gas is changed adiabatically from 5…

To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows:

$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$

Here, $P_{1}$ and $V_{1}$ are the initial pressure and volume, respectively, $P_{2}$ and $V_{2}$ are the final pressure and volume, respectively, and $\gamma$ is the heat capacity ratio of the gas.

Given in the question, $\gamma = 1.5$ , $V_{1} = 5$ litres and $V_{2} = 4$ litres. We need to find the ratio $\frac{P_{1}}{P_{2}}$ .

Rearranging the adiabatic equation for the ratio $\frac{P_{1}}{P_{2}}$ , we get:

$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma} \Rightarrow \frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma}$

Substituting the given values:

$\frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma} = \left(\frac{4}{5}\right)^{1.5}$

To calculate the value:

$\frac{P_{1}}{P_{2}} = \left(\frac{4}{5}\right)^{1.5} = \left(\frac{4}{5}\right)^{\frac{3}{2}}$

Simplifying further:

$\frac{P_{1}}{P_{2}} = \left(\frac{2^2}{5}\right)^{\frac{3}{2}} = \left(\frac{2^3}{5^{\frac{3}{2}}}\right) = \frac{8}{5\sqrt{5}}$

Therefore, the ratio of the initial pressure to the final pressure is $\frac{8}{5\sqrt{5}}$ , which corresponds to Option B.