Question

One mole of an ideal gas expands adiabatically at constant pressure such that its temperature $T \propto \frac{1}{\sqrt{V}}$. The value of $\gamma$ for the gas is $\left( \gamma = \frac{C_p}{C_v}, V = \text{Volume of the gas} \right)$

• A. 1.8
• B. 1.5
• C. 1.3
• D. 1.4

Hint:

Whenever a process relation is given in the form $T V^x = \text{constant}$, simply set $x = \gamma - 1$ to find gamma directly! Here, $x = 0.5$, so $\gamma = 0.5 + 1 = 1.5$. A clean and simple shortcut!

Answer 1

The Correct Option is B

Step 1: Recall the adiabatic relation.
For an adiabatic change, $TV^{\gamma-1}$ stays constant.

Step 2: Rewrite the given rule.
We are told $T\propto\frac{1}{\sqrt V}$, that is $T\propto V^{-1/2}$. Multiplying, $TV^{1/2}$ is constant.

Step 3: Match the powers of $V$.
Compare $TV^{\gamma-1}=$ const with $TV^{1/2}=$ const. So $\gamma-1=\frac{1}{2}$.

Step 4: Solve.
$\gamma=\frac{3}{2}=1.5$. \[ \boxed{1.5} \]