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Q10.15: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.15 Solution: Given below are observations on molar specific heats at room…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly apply the polyatomic gas formula ($C_p = \frac{9}{2}R$) to diatomic gases like $N_2$ and $O_2$ by assuming all degrees of freedom are active; however, at room temperature, vibrational modes are frozen, so only translational and rotational modes contribute, resulting in $C_p = \frac{7}{2}R \approx 29.1 \text{ J mol}^{-1}\text{K}^{-1}$.

Given below are observations on molar specific heats at room temperature of some common gases.

Step-by-Step Verified Solution

**Answer:** The molar specific heats of gases at room temperature are: Hydrogen (H_): 28.8 J mol__K__; Nitrogen (N_): 29.1 J mol__K__; Oxygen (O_): 29.4 J mol__K__; Carbon dioxide (CO_): 37.0 J mol__K__; Sulphur dioxide (SO_): 40.4 J mol__K__. **Step-by-Step Explanation:** For monatomic gases, Cv=32R=12.47 J mol−1K−1C_v = \frac{3}{2}R = 12.47 \text{ J mol}^{-1}\text{K}^{-1} and Cp=52R=20.78 J mol−1K−1C_p = \frac{5}{2}R = 20.78 \text{ J mol}^{-1}\text{K}^{-1}. For diatomic gases like H_, N_, O_, at room temperature, vibrational modes are not excited, so Cv=52R=20.78 J mol−1K−1C_v = \frac{5}{2}R = 20.78 \text{ J mol}^{-1}\text{K}^{-1} and Cp=72R=29.09 J mol−1K−1C_p = \frac{7}{2}R = 29.09 \text{ J mol}^{-1}\text{K}^{-1}. For polyatomic gases like CO_ and SO_, vibrational modes contribute, giving higher values. [citation:2]

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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