Question

The initial pressure and volume of a gas is 'P' and 'V' respectively. First by isothermal process gas is expanded to volume '9V' and then by adiabatic process its volume is compressed to 'V' then its final pressure is (Ratio of specific heat at constant pressure to constant volume = $\frac{3}{2}$)

• A. $6P$
• B. $27P$
• C. $3P$
• D. $9P$

Hint:

Combine both steps into a single scaling calculation! The net change in pressure for an isothermal expansion followed by an adiabatic compression back to the same initial volume can be written directly as $P_{\text{final}} = P \cdot \frac{1}{k} \cdot k^{\gamma} = P \cdot k^{\gamma - 1}$. Here, $k=9$ and $\gamma - 1 = 0.5$, so $P_{\text{final}} = P \cdot 9^{0.5} = 3P$.

Answer 1

The Correct Option is C

Step 1: Map out the two stages.
A gas starts at pressure $P$, volume $V$. First it expands isothermally to $9V$, then it is compressed adiabatically back to $V$. With $\gamma = \dfrac{3}{2}$, find the final pressure.
Step 2: Tools for each stage.
Isothermal: $PV = $ constant (Boyle's law). Adiabatic: $PV^{\gamma} = $ constant.
Step 3: Isothermal expansion ($V \to 9V$).
$P\cdot V = P_2 \cdot 9V$, so $P_2 = \dfrac{P}{9}$. The pressure drops as the gas expands.
Step 4: Set up the adiabatic compression ($9V \to V$).
$P_2(9V)^{\gamma} = P_3 (V)^{\gamma}$, so $P_3 = P_2\left(\dfrac{9V}{V}\right)^{\gamma} = P_2\,(9)^{\gamma}$.
Step 5: Evaluate $(9)^{3/2}$.
$(9)^{3/2} = (\sqrt{9})^3 = 3^3 = 27$.
Step 6: Combine.
$P_3 = \dfrac{P}{9}\times 27 = 3P$.
\[ \boxed{P_3 = 3P\ \text{(option 3)}} \]