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An Ideal Gas of Density 1.7 × 10−3 G Cm−3 at a Pressure of 1.5 × 105 Pa is Filled in a Kundt'S Tube. - Physics

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प्रश्न

An ideal gas of density 1.7 × 10−3 g cm−3 at a pressure of 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 formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.

संक्षेप में उत्तर
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उत्तर

Given:
Density of the ideal gas, ρ = 1.7 × 10−3 g/cm3
= 1.7 k/gm3
Pressure of the gas, P = 1.5 × 105 Pa
R = 8.3 J/mol-K
Resonance frequency of the gas = 3.0 kHz
Node separation in the Kundt's tube

`"l"/2 = 6 "cm"`

So, l = 2×6 = 12 cm = 12 × 10−2 m
So, V = fl = 3 × 103 × 12 × 10−2
          = 360 m/s
Speed of sound, V  =` sqrt( (gamma"p")/ρ)`

Or `"V"^2 =( gamma"p")/ρ `

`therefore gamma =("v"^2ρ)/"P" = ((360)^2 xx 1.7 xx 10^-3)/(1.5 xx 10^5)`

= 1.4688

Using `"C"_"P" -"C"_"v" = "R" and "C"_"p"/"C"_"v" = gamma`

We know that 

`"C"_"v" = "R"/(gamma-1) =8.3/0.4688`

= 17.7 J / mol -K

Cp = R +Cv =8.3 +17.7 = 26 J /mol -K

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Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
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अध्याय 5: Specific Heat Capacities of Gases - Exercises [पृष्ठ ८०]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 5 Specific Heat Capacities of Gases
Exercises | Q 34 | पृष्ठ ८०

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