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Subject: Physics , asked on 17/10/12

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

Dear Student!!

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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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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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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