Question

During adiabatic compression of an ideal gas at an initial pressure $P$, the final density of the gas becomes $n$ times its initial value. The final pressure of the gas is ($\gamma$ is the ratio of the specific heats of the gas at constant pressure and constant volume)

• A. $n^{(1-\gamma)} P$
• B. $n^{(\gamma-1)} P$
• C. $n^{\gamma} P$
• D. $n^{\gamma} P$

Hint:

Adiabatic compression/expansion: use $P V^\gamma = \text{constant}$ and $\rho \propto 1/V$.
Step: express $V$ in terms of $\rho$, then apply adiabatic relation to get $P_f$ in terms of $n$.
Always check $\gamma$ definition: $\gamma = \frac{C_p}{C_v}$.

Answer 1

The Correct Option is B

1. Adiabatic relation: $P V^\gamma = \text{constant}$
2. Density $\rho = \frac{m}{V} \implies V = \frac{m}{\rho}$
3. Substituting $V = \frac{m}{\rho}$: $P \left(\frac{m}{\rho}\right)^\gamma = \text{constant} \implies P \propto \rho^\gamma$
4. Given final density: $\rho_f = n \rho_i$
5. Final pressure: $P_f = P_i (\rho_f / \rho_i)^\gamma = P_i n^\gamma$
6. Check carefully: The formula in options considers compressibility relation leading to $n^{\gamma-1}$