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: Understanding the Question:
An ideal gas undergoes an adiabatic process where its thermodynamic temperature changes in inverse proportion to the square root of its volume. We need to determine the ratio of specific heats ($\gamma$) for this gas.

Step 2: Key Formula or Approach: The standard state relation governing temperature and volume during any reversible adiabatic process is: $$ T V^{\gamma - 1} = \text{constant} $$

Step 3: Detailed Explanation:
The question statement provides the temperature-volume proportionality rule: $$ T \propto \frac{1}{\sqrt{V}} \implies T \propto V^{-1/2} $$ We can convert this proportionality into an equation by introducing a constant $k$: $$ T = k \cdot V^{-1/2} $$ Multiplying both sides by $V^{1/2}$ to group the variables on the left: $$ T V^{1/2} = \text{constant} \quad \text{--- (Equation 1)} $$ Now, let's compare our experimental state equation with the standard theoretical adiabatic equation: $$ T V^{\gamma - 1} = \text{constant} \quad \text{--- (Equation 2)} $$ Equating the exponents of the volume parameter $V$ from Equation 1 and Equation 2: $$ \gamma - 1 = \frac{1}{2} $$ $$ \gamma = 1 + \frac{1}{2} = \frac{3}{2} = 1.5 $$

Step 4: Final Answer:
The value of $\gamma$ for the gas is 1.5, which corresponds to option (B).