Question

An ideal gas having pressure \(P\), volume \(V\) and temperature \(T\) is expanded isothermally, to a volume \(3V\) and final pressure \(P_I\). The same gas is expanded adiabatically to a volume \(3V\), the final pressure being \(P_A\). The ratio \(\frac{P_A}{P_I}\) is \((\frac{C_P}{C_V} = γ)\)

• A. \(3γ\)
• B. \(3^γ\)
• C. \(3^{(1-γ)}\)
• D. \(3^{(γ-1)}\)

Hint:

Isothermal: PV constant. Adiabatic: PV to the gamma constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In an isothermal process \(PV=\) constant. In an adiabatic process \(PV^\gamma=\) constant.

Step 2: Isothermal expansion
\[ PV=P_I(3V)\Rightarrow P_I=\frac P3 \]

Step 3: Adiabatic expansion
\[ PV^\gamma=P_A(3V)^\gamma\Rightarrow P_A=\frac{P}{3^\gamma} \]

Step 4: Ratio
\[ \frac{P_A}{P_I}=\frac{P/3^\gamma}{P/3}=3^{1-\gamma} \]

Step 5: Check
Since \(\gamma>1\), the exponent is negative and the ratio is below 1. This agrees with the fact that the adiabatic pressure falls more, because the gas also cools. The answer is \(3^{(1-\gamma)}\), option (C).

Final Answer:
The adiabatic pressure is P over 3 to the gamma and the isothermal one is P over 3, giving 3^(1 - gamma), option (C). \[ \boxed{3^{(1-\gamma)}} \]