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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

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)3⇒kTc=327ab

Substituting the above values in equation (1)

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

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Most popular questions from this chapter

By subtracting μNfrom localid="1648229964064" U,H,F,orG,one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),

Φ≡U-TS-μN.

(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives ofΦwith respect toT,V, and μN

(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φtends to decrease.

(c) Prove thatϕ=-PV.

(d) As a simple application, let the system be a single proton, which can be "occupied" either by a single electron (making a hydrogen atom, with energy -13.6eV) or by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800Kand an electron concentration of about 2×1019per cubic meter. Calculate Φfor both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in Problem 5.20, the prediction for such a small system is only a probabilistic one.)

Derive a formula, similar to equation 5.90, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.

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).

In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.

(a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written μA=μA°+kTln(1-x)where A is the chemical potential of pure A (at the same temperature and pressure) and x=NB/NA+NB. Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase.

(b) At any given temperature T, let x1 and xgbe the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that x1and xg obey the equations =1-xl1-xg=eΔGA°/RTandxlxg=eΔGB°/RT and where ΔG°represents the change in G for the pure substance undergoing the phase change at temperature T.

(c) Over a limited range of temperatures, we can often assume that the main temperature dependence of ΔG°=ΔH°-TΔS°comes from the explicit T; both ΔH°andΔS°are approximately constant. With this simplification, rewrite the results of part (b) entirely in terms of ΔHA°,ΔHB° TA, and TB (eliminating ΔGandΔS). Solve for x1and xgas functions of T.

(d) Plot your results for the nitrogen-oxygen system. The latent heats of the pure substances areΔHN2°=5570J/molandΔHO2°=6820J/mol. Compare to the experimental diagram, Figure 5.31.

(e) Show that you can account for the shape of Figure 5.32 with suitably chosenΔH° values. What are those values?

Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.
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