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QPage 76: Q10: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Page 76: Q10 Solution: The ratio for a gas is 1.29. What is the degree of freedom of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently plug $gamma =C_p/C_v$ into the ideal‑gas formula $\gamma=1+\frac{2}{f}$ and round the resulting non‑integer to the nearest whole number, ignoring that this relation assumes no active vibrational modes and a fixed integer $f$. In reality $gamma$ can reflect partial excitation of vibrational degrees, so a value like 1.29 does not guarantee a literal $f=7$.

The ratio Cp/CvC_p / C_v for a gas is 1.29. What is the degree of freedom of the molecules of this gas?

Step-by-Step Verified Solution

**Answer:** Degree of freedom f=7f = 7. **Step-by-Step Explanation:** γ=1+2f\gamma = 1 + \frac{2}{f}. Given γ=1.29\gamma = 1.29:
1.29=1+2f⇒0.29=2f⇒f=20.29≈6.9≈71.29 = 1 + \frac{2}{f} \Rightarrow 0.29 = \frac{2}{f} \Rightarrow f = \frac{2}{0.29} \approx 6.9 \approx 7
Note: In reality, no gas has more than 6 degrees of freedom; this value indicates complex molecular behaviour [citation:14].

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