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: Understanding the Question:
The problem compares two different thermodynamic processes starting from the exact same initial state $(P, V, T)$ to the same final volume $V' = 2V$.
In the first scenario, the gas undergoes an isothermal expansion to a final pressure $P_i$.
In the second scenario, the same gas undergoes an adiabatic expansion to a final pressure $P_a$. We need to find the ratio $\frac{P_i}{P_a}$.

Step 2: Key Formula or Approach:
1. For an isothermal process ($T = \text{constant}$):
$$PV = \text{constant} \implies P_1 V_1 = P_2 V_2$$ 2. For an adiabatic process:
$$PV^\gamma = \text{constant} \implies P_1 V_1^\gamma = P_2 V_2^\gamma$$

Step 3: Detailed Explanation:
Let's analyze the isothermal expansion first:
$$P \cdot V = P_i \cdot (2V) \implies P_i = \frac{P}{2}$$ Now, let's analyze the adiabatic expansion:
$$P \cdot V^\gamma = P_a \cdot (2V)^\gamma$$ $$P \cdot V^\gamma = P_a \cdot 2^\gamma \cdot V^\gamma$$ Canceling out $V^\gamma$ from both sides gives:
$$P_a = \frac{P}{2^\gamma}$$ Now, take the ratio of the final pressure of the isothermal expansion ($P_i$) to that of the adiabatic expansion ($P_a$):
$$\frac{P_i}{P_a} = \frac{\left(\frac{P}{2}\right)}{\left(\frac{P}{2^\gamma}\right)} = \frac{2^\gamma}{2} = 2^{\gamma - 1}$$

Step 4: Final Answer:
The ratio $\frac{P_i}{P_a}$ is equal to $2^{\gamma - 1}$, which matches option (B).