QJEEM24Ph8880: JEE Main 2024 Physics - Class 11 JEEM24Ph8880 Solution: During an adiabatic process, the pressure of a gas is found to…

For an adiabatic process, the following relation holds:

$P V^{\gamma} = \text{constant}$

where P is the pressure, V is the volume, and $\gamma = \frac{C_p}{C_v}$ .

We are given that the pressure is proportional to the cube of the absolute temperature:

$P \propto T^3$ .

Using the ideal gas law, $PV = nRT$ , we can rewrite this as:

$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$ .

Substituting this into the adiabatic relation, we get:

$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$ .

Simplifying, we have:

$P^{1-\gamma} T^{2\gamma} = \text{constant}$ .

Since P is proportional to $T^3$ , we can write:

$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$ .

This simplifies to:

$T^{3-3\gamma + 2\gamma} = \text{constant}$ .

For this equation to hold, the exponent of T must be zero. Therefore:

$3 - 3\gamma + 2\gamma = 0$ .

Solving for $\gamma$ , we get:

$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$ .

Therefore, the correct answer is Option B.