In Section 6.5 I derived the useful relation F=-kTln(Z)between the Helmholtz free energy and the ordinary partition function. Use analogous argument to prove that ϕ=-kT×ln(Z^), where Z^ is the grand partition function and ϕis the grand free energy introduced in Problem 5.23.

Short Answer

Expert verified

The grand canonical function is ϕ=-kT×ln(Z^) which is proved successfully.

Step by step solution

01

Step 1. Grand Potential differentiate

Formula for grand potential is:

ϕ=U-TS-μN

ϕ=U-TS-μN

where, Sis entropy, Tis temperature, Nis particles number and μis Gibbs free energy per particle.

Use thermodynamic equation:

dU=TdS-PdV+μdN

Equation for an infinitesimal change in grand potential is:

dϕ=d(U-TS-μN)=dU-d(TS)-d(μN)=dU-TdS-SdT-μdN-Ndμ

Substitute dU=TdS-PdV+μdN

localid="1647239388505" dϕ=TdS-PdV+μdN-TdS-SdT-μdN-Ndμ=-SdT-PdV-Ndμ

Differentiate equation dϕ=-SdT-PdV-Ndμwith respect to μ

ϕμTV=-N

02

Step 2. Number of particles equation

We differentiate function ϕ~=-kT×ln(Z)^ with respect to μ.

ϕ~μTV=-kTZ^Z^μ

Here, Z^ is grand partition function.

As, average number of particles is given by:

localid="1647240173822" N¯=kTZ^Z^μ

Substitute kTZ^Z^μ=N¯ in above differential equation,

ϕ~μTV=-N¯

03

Step 3. Canonical function equation

For grand canonical function ϕandϕ~at μ=0we get,

ϕ~=-kT×ln(Z)=F

Here, F=-kT×ln(Z)is called Helmholtz free energy.

Substitute μ=0in equation ϕ=U-TS-μN

ϕ=U-TS

As, Helmholtz free energy is F=U-TS

Substitute F=U-TSin equation ϕ=U-TSwe get,

ϕ=F

So, ϕand ϕ~at same initial conditions have same values. Therefore, they are the same functions. Hence, the grand canonical function is ϕ=-kT×ln(Z^).

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Most popular questions from this chapter

Number of photons in a photon gas.

(a) Show that the number of photons in equilibrium in a box of volume V at temperature T is

N=8πVkThc30x2ex-1dx

The integral cannot be done analytically; either look it up in a table or evaluate it numerically.

(b) How does this result compare to the formula derived in the text for the entropy of a photon gas? (What is the entropy per photon, in terms of k?)

(c) Calculate the number of photons per cubic meter at the following temperatures: 300 K; 1500 K (a typical kiln); 2.73 K (the cosmic background radiation).

The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they "heated" the photon radiation but not the neutrino radiation.

(a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the auxiliary functions u(T) and f(T) introduced in the previous problem. Argue that this total entropy would have ben conserved in the early universe, assuming that no other species of particles interacted with these.

(b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by

TTν32π445+u(T)+f(T)=constant

as the universe expands and cools. Evaluate the constant by assuming that T=Tv when the temperatures are very high.

(c) Calculate the ratio T/Tv, in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K.

(d) Use a computer to plot the ratio T/Tv, as a function of T, for kT/mc2ranging from 0 to 3.*

The Sommerfeld expansion is an expansion in powers of kTεF, which is assumed to be small. In this section I kept all terms through order kTεF2, omitting higher-order terms. Show at each relevant step that the term proportional to localid="1650117451748" T3is zero, so that the next nonvanishing terms in the expansions forlocalid="1650117470867" μand localid="1650117476821" Uare proportional to localid="1650117458596" T4. (If you enjoy such things, you might try evaluating the localid="1650117464980" T4terms, possibly with the aid of a computer algebra program.)

The planet Venus is different from the earth in several respects. First, it is only 70% as far from the sun. Second, its thick clouds reflect 77%of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light.

(a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.

(b) Estimate the surface temperature again, taking the reflectivity of the clouds into account.

(c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly 70times that of earth's atmosphere. You can therefore model the atmosphere of Venus as 70successive "blankets" of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of 21/4. The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)

Evaluate the integrand in equation 7.112as a power series in x, keeping terms through x4• Then carry out the integral to find a more accurate expression for the energy in the high-temperature limit. Differentiate this expression to obtain the heat capacity, and use the result to estimate the percent deviation of Cvfrom3NkatT=TDandT=2TD.

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