During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of $\frac{\mathrm{Cp}}{\mathrm{Cv}}$ for the gas is :

<p>For an adiabatic process, the following relation holds:</p> <p>$P V^{\gamma} = \text{constant}$</p> <p>where P is the pressure, V is the volume, and $\gamma = \frac{C_p}{C_v}$.</p> <p>We are given that the pressure is proportional to the cube of the absolute temperature: </p> <p>$P \propto T^3$.</p> <p>Using the ideal gas law, $PV = nRT$, we can rewrite this as:</p> <p>$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$.</p> <p>Substituting this into the adiabatic relation, we get:</p> <p>$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$.</p> <p>Simplifying, we have:</p> <p>$P^{1-\gamma} T^{2\gamma} = \text{constant}$.</p> <p>Since P is proportional to $T^3$, we can write:</p> <p>$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$.</p> <p>This simplifies to:</p> <p>$T^{3-3\gamma + 2\gamma} = \text{constant}$.</p> <p>For this equation to hold, the exponent of T must be zero. Therefore:</p> <p>$3 - 3\gamma + 2\gamma = 0$.</p> <p>Solving for $\gamma$, we get:</p> <p>$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$.</p> <p>Therefore, the correct answer is <strong>Option B</strong>. </p>