during an adiabatic process, the pressure of a gas is found to be proportional to the cube of its…

Explanation:

Step1: Write adiabatic relation

For an adiabatic process, $PV^{\gamma}=C$. Also, from ideal - gas law $PV = nRT\Rightarrow V=\frac{nRT}{P}$. Substituting $V$ in adiabatic equation, we get $P(\frac{nRT}{P})^{\gamma}=C$. Simplifying, $P^{1 - \gamma}T^{\gamma}=C$.

Step2: Use given proportionality

Given $P\propto T^{3}$, so $P = kT^{3}$ (where $k$ is a constant). Comparing with $P^{1 - \gamma}T^{\gamma}=C$, we have $P = C^{\frac{1}{1 - \gamma}}T^{\frac{\gamma}{1 - \gamma}}$. Then $\frac{\gamma}{1 - \gamma}=3$.

Step3: Solve for $\gamma$

Cross - multiply the equation $\frac{\gamma}{1 - \gamma}=3$: [ \begin{align*} \gamma&=3(1 - \gamma)\ \gamma&=3-3\gamma\ 4\gamma&=3\ \gamma&=\frac{3}{2} \end{align*} ] Since $\gamma=\frac{C_{P}}{C_{V}}$, the ratio $\frac{C_{P}}{C_{V}}=\frac{3}{2}$.

Answer:

1. 3/2