Question

A monatomic gas of volume \( V \) and pressure \( P \) expands isothermally to a volume \( 27V \) and then compressed adiabatically to a volume \( V \). The final pressure of the gas is:

• A. \( 3P \)
• B. \( 2P \)
• C. \( 9P \)
• D. \( 4P \)

Hint:

In an adiabatic process, the pressure and volume are related by \( P V^\gamma = \text{constant} \) where \( \gamma \) is the specific heat ratio for the gas.

Answer 1

The Correct Option is C

For an isothermal expansion, we know that the pressure and volume are inversely proportional: \[ P_1 V_1 = P_2 V_2 \] Given that the initial volume is \( V \) and the final volume is \( 27V \), the initial and final pressures for the isothermal process are: \[ P_1 V = P_2 \cdot 27V \quad \Rightarrow \quad P_2 = \frac{P}{27} \] For the subsequent adiabatic compression, we use the formula for adiabatic processes: \[ P V^{\gamma} = \text{constant} \] For a monatomic gas, \( \gamma = \frac{5}{3} \). Applying this to the isothermal pressure and volume: \[ P_2 V^{\gamma} = P_3 V^{\gamma} \quad \Rightarrow \quad \left( \frac{P}{27} \right) \cdot (27V)^{\gamma} = P_3 \cdot V^{\gamma} \] Simplifying the equation: \[ P_3 = 9P \] Thus, the final pressure is \( 9P \).

Answer 2

Step 1: Understanding the process
The gas undergoes two processes:
1. Isothermal expansion from volume \( V \) to \( 27V \) at constant temperature.
2. Adiabatic compression from volume \( 27V \) back to \( V \).

Step 2: Isothermal expansion
For isothermal process, \( P V = \text{constant} \), so:
\[ P \times V = P_1 \times 27V \implies P_1 = \frac{P}{27} \]
where \( P_1 \) is the pressure after expansion.

Step 3: Adiabatic compression
For adiabatic process, the relation is:
\[ P V^\gamma = \text{constant} \]
where \( \gamma = \frac{C_p}{C_v} \). For monatomic gas, \( \gamma = \frac{5}{3} \).

Using initial conditions for adiabatic process:
\[ P_1 (27V)^\gamma = P_2 (V)^\gamma \]
Substitute \( P_1 = \frac{P}{27} \):
\[ \frac{P}{27} \times (27V)^\gamma = P_2 \times V^\gamma \]

Step 4: Simplify the expression
\[ \frac{P}{27} \times 27^\gamma V^\gamma = P_2 V^\gamma \]
Divide both sides by \( V^\gamma \):
\[ \frac{P}{27} \times 27^\gamma = P_2 \]
Since \( 27 = 3^3 \),
\[ 27^\gamma = (3^3)^\gamma = 3^{3\gamma} = 3^{3 \times \frac{5}{3}} = 3^{5} = 243 \]
Therefore:
\[ P_2 = \frac{P}{27} \times 243 = 9P \]

Step 5: Conclusion
The final pressure of the gas after the two processes is \( 9P \).