Pls explain the relation between the ratio Cp/Cv and atomicity of gas in detail.

Asked by Anvita Chaudhary | 16th Dec, 2013, 07:32: PM

Expert Answer:

The internal energy associated with any molecule depends upon its degrees of freedom. It means different modes of translational, rotational and vibrational motion. Monoatomic gases like He or Ar etc. have only translational degrees of freedom. A polyatomic molecule consisting of N atoms has following degrees of freedom:

Linear molecules:

Translational – 3, Rotational – 2 and Vibrational – 3 N-5

 

Non-linear molecules:

Translational – 3, Rotational – 3 and Vibrational – 3 N-6

 

Each translational and rotational degree of freedom contributes energy = ½ kT. Each vibrational degree of freedom contributes energy = kT.  Vibrational degrees of freedom contribute only at high temperature. They do not contribute at ordinary temperatures.

 

Thus, for monoatomic gases, energy (U) per mole = 3*(1/2)RT= 3/2RT

Therefore,

Cv = (dU/dT)v = 3/2 R

Diatomic gases such as H2 have not only translational motion but also have rotational and vibrational motion. At ordinary temperature, contribution by vibrational motion is negligible but it has two rotational degrees of motion. Hence, total energy (U) per mole of a diatomic molecule = (3*1/2RT)+(2*1/2RT)=(5/2RT)

Therefore,

Cv for diatomic gas = 5/2R

As Cp=Cv+R, henceCv=7/2R and Cp = 9/2 R

Hence the ratio Cp/Cv  will be:

For monoatomic gas such as He and Ar:

(5/2)R/(3/2)R= 5/3=1.66

For diatomic gas (H2, O2 or CO):

(7/2)R/(5/2)R=7/5=1.40

For triatomic gas such as (CO2):

(9/2)R/(7/2)R=9/7=1.30

 

 

Answered by Prachi Sawant | 20th Dec, 2013, 02:55: PM

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