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

δPδV=-NkT(V-Nb)2+2aN2V3 (2)

Again differentiating we get

δ2Pδ2V=NkT(V-Nb)3-6aN2V4 (3)

02

At critical point

δPδV=0δ2Pδ2V=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)3kTc=327ab

Substituting the above values in equation (1)

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

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Most popular questions from this chapter

The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:

NaAlSi3O8NaAlSi2O6+SiO2

Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both S and V.

Assume that the air you exhale is at 35°C, with a relative humidity of 90%. This air immediately mixes with environmental air at 5°C and unknown relative humidity; during the mixing, a variety of intermediate temperatures and water vapour percentages temporarily occur. If you are able to "see your breath" due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapour pressure graph drawn in Problem 5.42.)

How can diamond ever be more stable than graphite, when it has

less entropy? Explain how at high pressures the conversion of graphite to diamond

can increase the total entropy of the carbon plus its environment.

Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 x 10-6 bar-1, while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

Use the result of the previous problem to calculate the freezing temperature of seawater.

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