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: Note what is given.
An adiabatic process follows $PV^{3/2}=\text{constant}$. The starting temperature is $T$, and the volume is squeezed to half. We want the final temperature.

Step 2: Bring in the gas law.
The ideal gas law $PV=nRT$ tells us $P=\frac{nRT}{V}$, so $P$ is proportional to $\frac{T}{V}$. We use this to swap $P$ for $T$ and $V$.

Step 3: Substitute into the given relation.
Put $P\propto\frac{T}{V}$ into $PV^{3/2}=\text{constant}$: \[ \frac{T}{V}\cdot V^{3/2}=\text{constant}. \]

Step 4: Simplify the powers of $V$.
\[ T\cdot V^{-1}\cdot V^{3/2}=T\cdot V^{1/2}=\text{constant}. \] So the rule connecting two states is $T_1V_1^{1/2}=T_2V_2^{1/2}$.

Step 5: Put in the values.
$T_1=T$, $V_1=V$, and $V_2=\frac{V}{2}$: \[ T\cdot V^{1/2}=T_2\left(\frac{V}{2}\right)^{1/2}. \]

Step 6: Solve for $T_2$.
Divide both sides by $V^{1/2}$: \[ T=T_2\left(\frac{1}{2}\right)^{1/2}=\frac{T_2}{\sqrt2}, \] so $T_2=\sqrt2\,T$.

Step 7: State the result.
The final temperature is $\sqrt2\,T$, which is option (3).
\[ \boxed{\sqrt2\,T} \]