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Chapter 5: Q. 5.48 (page 185)

As you can see from Figure5.20,5.20,the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives ofPPwith respect toVV(at fixedTT) are zero. Use this fact to show that

Vc=3Nb, Pc =127ab2 and kTc=827ab

Short Answer

Expert verified

Vc=3Nb, Pc=127ab2and kTc=827ab

Step by step solution

01

van der Waal's equation

P=NkT(V-Nb)-aN2V2 (1)

Partial differentiation of the equation w.r.t V

we get

未P未V=-NkT(V-Nb)2+2aN2V3 (2)

Again differentiating we get

未2P未2V=NkT(V-Nb)3-6aN2V4 (3)

02

At critical point

未P未V=0未2P未2V=0

NkTc(Vc-Nb)2=2aN2Vc3andNkTc(Vc-Nb)3=6aN2Vc4

03

Step 3:  Finding Vc, Tc, Pc.

On equating the above equations we get

Vc=3Nb (4)

Substituting the equation (4) in equation (2)

NkTc(3Nb-Nb)2=2aN2(3Nb)3鈬�kTc=327ab

Substituting the above values in equation (1)

Pc=Nk(3a27b)(3Nb-Nb)-aN2(3Nb)2鈬�Pc=127ab2

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Most popular questions from this chapter

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