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`         An ideal gas having density 1.7 × 10-3 g/cm3 at a pressure 1.5 × 105 Pa is filled in a Kundt’s tube. When the gas is resonated at a frequency of 3.0 kHz. Nodes are heat capacities Cp and Cv of the gas.`
5 years ago

```							Sol. Given fo = 1.7 ×10^–3 g/cm^3 = 1.7 kg/m^3, P = 1.5 × 10^5 Pa, R = 8.3 J/mol-k,
f = 3.0 KHz.
Node separation in a Kundt’` tube = λ/2 = 6 cm, ⇒ λ = 12 cm = 12  * 10^3 m
So, V = f λ = 3 * 10^3 * 12 * 10^-2 = 360 m/s
We know, Speed of sound = √γP/fo ⇒ (360)^2 = γ * 1.5 * 10^5/1.7 ⇒ γ = (360)^2 * 1.7/1.5 * 10^5 = 1.4688
But C base V = R/γ – 1 = 8.3/1.488 – 1 = 17.72 J/mol-k
Again C base P/C base V = γ So, C base P = γC base V = 17.72 * 1.468 = 26.01 = 26 J/mol-k

```
5 years ago
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