Physics, asked by ANANDIKSHA935, 1 year ago

Relation between kinetic energy and degree of freedom

Answers

Answered by sssrohit005p4c0ey
0
First of all, if we have an ideal gas then pV=nRTpV=nRT, so p=(nRT)/Vp=(nRT)/V. Then, W=(nRT)/VdVW=(nRT)/VdV. Your Work equation is wrong.

To clarify on how to develop from dU=dWdU=dW in an alternative way, we can obtain some interesting properties.

We have dU=dW→U=WdU=dW→U=W , for an adiabatic process. (It is also known that Cp=(f/2)nR+nRCp=(f/2)nR+nR and Cv=(f/2)nRCv=(f/2)nR ; Heat capacities for constant pressure and volume, respectively)

So:

dU=(f/2)nRdT=(Cv)dT,asCv=(f/2)nRdU=(f/2)nRdT=(Cv)dT,asCv=(f/2)nR

Now we can develop the equation:

(Cv)dT=−((nRT)/V)dV<−>(Cv)dT=−((nRT)/V)dV<−>

(Cv)dT/T=−((Cp−Cv)/V)dV(Cv)dT/T=−((Cp−Cv)/V)dV

Integrating both terms, (assuming T0T0,V0V0as initial TT and VV, respectively)

Cv∗ln(T/T0)=−(Cp−Cv)∗ln(V/V0)<−>Cv∗ln(T/T0)=−(Cp−Cv)∗ln(V/V0)<−>

ln((T/T0)Cv)=ln((V/V0)(Cv−Cp))<−>ln((T/T0)Cv)=ln((V/V0)(Cv−Cp))<−>

T/T0=(V/V0)((Cv−Cp)/Cv)T/T0=(V/V0)((Cv−Cp)/Cv)

And as γ=(Cv/Cp)γ=(Cv/Cp): T/T0=(V/V0)(1−γ)T/T0=(V/V0)(1−γ) ; where γγ is a greek letter, which represents the adiabatic constant.

We can say that, TV(γ−1)=T0V0=constantTV(γ−1)=T0V0=constant We can also obtain PVγPVγ = constant applying the Ideal gas law in TV(γ−1)=constantTV(γ−1)=constant.

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