Question

An ideal gas with pressure $P$, volume $V$ and temperature $T$ is expanded isothermally to a volume $2V$ and a final pressure $P_i$. The same gas is expanded adiabatically to a volume $2V$, the final pressure is $P_a$. In terms of the ratio of the two specific heats for the gas $\gamma$, the ratio $\frac{P_i}{P_a}$ is

• A. $2^{\gamma + 1}$
• B. $2^{\gamma - 1}$
• C. $2^{1 - \gamma}$
• D. $2^\gamma$

Hint:

An adiabatic curve is steeper than an isothermal curve on a $P\text{-}V$ indicator diagram. Since the pressure drops more dramatically during an adiabatic expansion than during an isothermal expansion to the same volume, $P_i > P_a$ must be true. Because $\gamma > 1$ for all real gases, the exponent $(\gamma - 1)$ is strictly positive, giving a ratio greater than 1.

Answer 1

The Correct Option is B

Step 1: Set the common start.
Both processes begin at the same state $(P, V, T)$ and end at the same volume $2V$.
Step 2: Apply the isothermal rule.
For constant temperature $PV$ is constant: \[ PV = P_i(2V) \;\Rightarrow\; P_i = \frac{P}{2}. \]
Step 3: Apply the adiabatic rule.
Here $PV^\gamma$ is constant: \[ PV^\gamma = P_a(2V)^\gamma. \]
Step 4: Solve for $P_a$.
Cancel $V^\gamma$ from both sides: \[ P_a = \frac{P}{2^\gamma}. \]
Step 5: Take the ratio.
\[ \frac{P_i}{P_a} = \frac{P/2}{P/2^\gamma} = \frac{2^\gamma}{2}. \]
Step 6: Simplify the power.
\[ \frac{P_i}{P_a} = 2^{\gamma - 1}. \] Since $\gamma > 1$, the isothermal pressure is the larger one, which makes physical sense. This is option (B). \[ \boxed{\frac{P_i}{P_a} = 2^{\gamma - 1}} \]