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i want derivation of cp and cv for monoatomic, diatomic, triatomic..... plz answer fast its urgent 2morrow is my exam..

Dear Student!!

Please refer the answer posted by Aman. 

Similarly you can for diatomic and tri atomic gas gases.

Diatomic gas has 5 degrees of freedom and tri atomic gas has 6 degrees of freedom.  

@ Aman - Good effort keep it up!!

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By definition of Specific heat

  Cv =dQ/(µ+dT)

by first law of thermodynamics

  dQ=dU+dW

at constant volume (dW=0) then

  dQ=dU

But U=fµRT/2, so that

dU/dT=fµR/2

Then molar specific heat at constant volume,

  Cv=dU/µdT

Here

  Cv=fR/2

Here  Cp-Cv=R

then

  Cp=Cv+R

Hence

  Cp=(f/2+1)R

now, rate of molar specific heat,

Cp/Cv=Γ  (Γ is gamma)

then

Γ=[(f/2+1)R]/(f/2)r

█ For monoatomic gas (f=3)

Cv=3R/2

Cp=5R/2

Γ=5/3

█ For diatomic gas  (f=5)

Cv=5R/2

Cp=7R/2

Γ=7/5

█ For Triatomic gas

▄Linear Structural  (f=7)

Cv=7R/2

Cp=9R/2

Γ=9/7

▄non-linear structural  (f=6)

Cv=3R

Cp=4R

Γ=4/3

Hence Proved.

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thanxxx  sir

  • -2

For Monoatomic gases

No of degrees are 3.All Three are translational energies

Hence the total internal energy for 1 mole of gas is

U=(1/3KBT+. 1/3KBT+1/3KBT)NA               (Acc to law of equipartition law)(K=boltzman constant)

U=3/2KBTNA  

diff. it both sides w.r.t. T

=dU/dT=3/2R                         (R=KBNA)

but dU/dT=Cv

therefore Cv =3/2R

hence Cp=CV+R

Cp=3/2R+R

CP=5/2R

Similarly you can solve for Diatomic gases but it has 5 degree of freedom.

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