Question

For an ideal gas, if the ratio of molar specific heats \(\gamma = 1.4\), then the specific heat at constant pressure \(C_p\), specific heat at constant volume \(C_v\) and corresponding molecule are respectively

• A. \(\dfrac{5}{2}R,\ \dfrac{3}{2}R,\) monoatomic.
• B. \(\dfrac{9}{2}R,\ \dfrac{7}{2}R,\) polyatomic.
• C. \(\dfrac{7}{2}R,\ \dfrac{5}{2}R,\) non-rigid diatomic.
• D. \(\dfrac{7}{2}R,\ \dfrac{5}{2}R,\) rigid diatomic.

Hint:

For rigid diatomic gases, \(\gamma = 1.4\). Monoatomic gases have \(\gamma = 1.67\).

Answer 1

The Correct Option is D

Step 1: Use the definition of \(\gamma\).
\[ \gamma = \frac{C_p}{C_v} \]
Step 2: Substitute given value.
\[ \frac{C_p}{C_v} = 1.4 = \frac{7}{5} \]
Step 3: Identify type of gas.
For a rigid diatomic gas:
\[ C_v = \frac{5}{2}R \quad \text{and} \quad C_p = C_v + R = \frac{7}{2}R \]
Step 4: Verify ratio.
\[ \gamma = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} = 1.4 \]
Step 5: Conclusion.
The gas is rigid diatomic with \(C_p = \dfrac{7}{2}R\) and \(C_v = \dfrac{5}{2}R\).