Question

The relation obeyed by a perfect gas during an adiabatic process is $PV^{3/2} = \text{constant}$. The initial temperature of the gas is $T$. When the gas is compressed to half of its initial volume, the final temperature of the gas is

• A. $2\sqrt{2}T$
• B. $4T$
• C. $\sqrt{2}T$
• D. $2T$

Hint:

Always convert $P-V$ relations to $T-V$ relations when the question asks about temperature changes. Substituting $P \propto T/V$ into $PV^\gamma = \text{const}$ instantly yields $TV^{\gamma-1} = \text{const}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a specific adiabatic relationship between pressure and volume for an ideal gas. We must find the final temperature when the volume is halved.

Step 2: Key Formula or Approach:
The given relation is $PV^{3/2} = \text{constant}$.
To relate volume and temperature, use the ideal gas equation $PV = nRT$, which implies $P$ is proportional to $\frac{T}{V}$.
Substitute $P \propto \frac{T}{V}$ into the given adiabatic relation to establish a direct relationship between $T$ and $V$.

Step 3: Detailed Explanation:
Starting with the given equation:
$$P \cdot V^{3/2} = \text{constant}$$ Replace $P$ with $\frac{T}{V}$ (ignoring $nR$ as it just merges into the constant):
$$\left(\frac{T}{V}\right) \cdot V^{3/2} = \text{constant}$$ $$T \cdot V^{-1} \cdot V^{3/2} = \text{constant}$$ $$T \cdot V^{1/2} = \text{constant}$$ This gives us the state equation:
$$T_1 V_1^{1/2} = T_2 V_2^{1/2}$$ Let the initial state be $T_1 = T$ and $V_1 = V$.
The final volume is halved, so $V_2 = \frac{V}{2}$.
Substitute these into the state equation:
$$T \cdot (V)^{1/2} = T_2 \cdot \left(\frac{V}{2}\right)^{1/2}$$ Divide both sides by $V^{1/2}$:
$$T = T_2 \cdot \left(\frac{1}{2}\right)^{1/2}$$ $$T = \frac{T_2}{\sqrt{2}}$$ Isolate the final temperature $T_2$:
$$T_2 = \sqrt{2}T$$

Step 4: Final Answer: The final temperature of the gas is $\sqrt{2}T$, matching option (C).