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: Understanding the Question:
An ideal gas undergoes a two-step thermodynamic process: first, an isothermal expansion from volume $V$ to $9V$, followed by an adiabatic compression returning the volume back to $V$. We need to calculate the final pressure of the gas at the end of these two steps.

Step 2: Key Formula or Approach:
1.

For the Isothermal Stage: Temperature is constant, satisfying Boyle's Law:
$$P_1 V_1 = P_2 V_2$$ 2.

For the Adiabatic Stage: The system follows Poisson's relation:
$$P_2 V_2^{\gamma} = P_3 V_3^{\gamma}$$ where $\gamma = \frac{C_p}{C_v} = \frac{3}{2} = 1.5$.

Step 3: Detailed Explanation:
Let's track the state parameters across both processes:

Initial State 1: Pressure $= P$, Volume $= V$

Step 1: Isothermal Expansion to State 2 ($V_2 = 9V$): $$P \cdot V = P_2 \cdot (9V) \implies P_2 = \frac{P}{9}$$

Step 2: Adiabatic Compression to State 3 ($V_3 = V$): Using the adiabatic equation between state 2 and state 3:
$$P_2 V_2^{\gamma} = P_3 V_3^{\gamma} \implies P_3 = P_2 \left(\frac{V_2}{V_3}\right)^{\gamma}$$ Substitute the volume values and $P_2 = \frac{P}{9}$ into the expression:
$$P_3 = \left(\frac{P}{9}\right) \cdot \left(\frac{9V}{V}\right)^{3/2}$$ $$P_3 = \frac{P}{9} \cdot (9)^{3/2}$$ To compute $(9)^{3/2}$ easily, take the square root of $9$ first ($\sqrt{9} = 3$), and then cube it ($3^3 = 27$):
$$P_3 = \frac{P}{9} \cdot 27 = 3P$$

Step 4: Final Answer:
The final pressure of the gas is $3P$, which corresponds to option (C).