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Chapter 9: Q23E (page 403)

The entropy of an ideal monatomic gas is(3/2)NkBlnE+,NkBlnV-NkBlnN to within an additive constant. Show that this implies the correct relationship between internal energy Eand temperature.

Short Answer

Expert verified

The expression for the internal energy of an ideal monatomic gas is in agreement with the kinetic theory of gases, and thus the correct relationship between E and T.

E=32NkBT

Step by step solution

01

Concept used

Equation of relationship between the internal energy and temperature for the entropy is

∂S∂E=1T

02

Put the given entropy in the equation

Assuming constant volume V for the ideal monatomic gas, the temperature T relates to the energy rate of change E of entropy S according to the equation:

∂S∂E=1T

Taking the partial derivative of our expression for S with respect to E we obtain:

∂S∂E=∂∂E32NkBlnE+NkBlnV-NkBlnN=32NkB1E=32NkBE

∂S∂E=32NkBE=1T

Rearranging for E we thus obtain:

32NkBE=1TE=32NkBT

This expression for the internal energy of an ideal monatomic gas is in agreement with the kinetic theory of gases, and thus the correct relationship between E and T.

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Most popular questions from this chapter

The electrons’ contribution to the heat capacity of a metal is small and goes to 0as T→0. We might try to calculate it via the total internal energy, localid="1660131882505" U=∫EN(E)D(E)dE, but it is one of those integrals impossible to do in closed form, and localid="1660131274621" N(E)FDis the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value.

(a) Starting withN(E)FDexpressed as in equation (34), show that the slope N(E)FDdEatE=EFis-1(4kBT).

(b) Based on part (a), the accompanying figure is a good approximation to N(E)FDwhen Tis small. In a normal gas, such as air, whenTis raised a little, all molecules, on average, gain a little energy, proportional to kBT. Thus, the internal energy Uincreases linearly with T, and the heat capacity, ∂U∂T, is roughly constant. Argue on the basis of the figure that in this fermion gas, as the temperature increases from 0to a small value T, while some particles gain energy of roughly kBT, not all do, and the number doing so is also roughly proportional to localid="1660131824460" T. What effect does this have on the heat capacity?

(c)Viewing the total energy increase as simply ∆U= (number of particles whose energy increases) (energy change per particle) and assuming the density of states is simply a constant Dover the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to kBREFT. (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore).

Calculate the Fermi energy for copper, which has a density of8.9×103kg/m3and one conduction electron per atom. Is room temperature "cold"?

By carrying out the integration suggested just before equation (9-28), show that the average energy of a one-dimensional oscillator in the limit kBT≫hω0iskBT.

Heat capacity (at constant volume) is defined as∂U/∂T. (a) Using a result derived in Example 9.6. obtain an expression for the heat capacity per unit volume, inJ/K⋅mi3, of a photon gas. (b) What is its value at300K?

Consider the two-sided room, (a) Which is more likely to have an imbalance of five particles (i.e. Nk=12N+S,S): a room withN=20or a room withrole="math" localid="1658330090284" N=60? (Note: The total number of ways of distributing particles. the sum ofWNhNfrom 0 toN, is2N.) (b) Which is more likely to have an imbalance of5%(i.e. NR=12N+0.05N,)? (c) An average-size room is quite likely to have a trillion mote air molecules on one side than on the other, what may we say that precisely half will be on each side?
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