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QPages 77-80: Q35: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q35 Solution: Standing waves of frequency 5.0 kHz are produced in a tube…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the distance between consecutive nodes as the full wavelength, forgetting that node‑to‑node spacing equals $\lambda/2$. Using $\lambda$ = node separation halves the sound speed and leads to incorrect $C_p$ and $C_v$ values; always double the node spacing to obtain $\lambda$ and keep units (kg vs g) consistent when applying $v=\sqrt{\gamma RT/M}$.

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities

C_p

and

C_v

of the gas.

Step-by-Step Verified Solution

**Answer:**

C_p = 1.0

J/g-K,

C_v = 0.71

J/g-K. **Step-BY-Step Explanation:** Given: O_ at 300 K,

f = 5.0

kHz, node separation = 3.3 cm.

\lambda = 2 \times 3.3 = 6.6

cm = 0.066 m.

v = 5000 \times 0.066 = 330

m/s. For oxygen,

M = 0.032

kg/mol,

\gamma = 1.4

.

v = \sqrt{\gamma RT/M} \Rightarrow T = \frac{v^2M}{\gamma R} = \frac{330^2 \times 0.032}{1.4 \times 8.314} = 300

K (consistent).

C_v = \frac{R}{M(\gamma-1)} = \frac{8.314}{0.032 \times 0.4} = 649

J/kg-K = 0.65 J/g-K.

C_p = \gamma C_v = 1.4 \times 0.65 = 0.91

J/g-K.

All Questions in this Chapter

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