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# Consider a gas of non interacting spin-0 bosons at high

ISBN: 9780201380279 40

## Solution for problem 75P Chapter 7

An Introduction to Thermal Physics | 1st Edition

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Problem 75P

Problem 75P

Consider a gas of non interacting spin-0 bosons at high temperatures, when T ≫ Tc. (Note that “high” in this sense can still mean below 1 K.)

(a) Show that, in this limit, the Bose-Einstein distribution function can be written approximately as

(b) Keeping only the terms shown above, plug this result into equation to derive the first quantum correction to the chemical potential for a gas of bosons.

(c) Use the properties of the grand free energy (Problems 1 and 2) to show that the pressure of any system is given by P = (kT/V) In Z, where Z is the grand partition function. Argue that, for a gas of non interacting particles, In z can be computed as the sum over all modes (or single-particle states) of In zi, where Zi is the grand partition function for the ith mode.

(d) Continuing with the result of part (c), write the sum over modes as an integral over energy, using the density of states. Evaluate this integral explicitly for a gas of noninteracting bosons in the hightemperature limit, using the result of part (b) for the chemical potential and expanding the logarithm as appropriate. When the smoke clears, you should find

again neglecting higher-order terms. Thus, quantam statistics results in a lowering of the pressure of a boson gas, as one might expect.

(e) Write the result of part (d) in the form of the virial expansion introduced in Problem 3, and read off the second virial coefficient, B(T). Plot the predicted B(T) for a hypothetical gas of noninteracting helium-4 atoms.

(f) Repeat this entire problem for a gas of spin-1/2 ferrnions. (Very few modifications are necessary.) Discuss the results, and plot the predicted virial coefficient for a hypothetical gas of noninteracting helium-3 atoms.

Problem 1:

By subtracting μN from U, H, F, or G, 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 of Φ with respect to T, V, and μ.

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

(c) Prove that Φ = – PV.

(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of 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 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for 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 below Problem 4, the prediction for such a small system is only a probabilistic one.)

Problem 4:

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)

Problem 2:

In Section 6.5, I derived the useful relation F =  kT ln Z between the Helmholtz free energy and the ordinary partition function. Use an analogous argument to prove that

ϕ =  kTlnZ

where z is the grand partition function and ϕ is The grand free energy introduced in Problem 5.

Problem 5:

By subtracting μN from U, H, F, or G, 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 of Φ with respect to T, V, and μ.

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

(c) Prove that Φ = – PV.

(d) As a simple application, let the system he a single proton, which can be “occupied” either by a single electron (making a hydrogen atom, with energy –13.6 eV) of 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 5800 K and an election concentration of about 2 × 1019 per cubic meter. Calculate Φ for 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 below Problem 6, the prediction for such a small system is only a probabilistic one.)

Problem 6:

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.)

Problem 3:

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:

(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)

(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

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1/11/16 Atomic Structure JJ Thompson • Determined the charge-to-mass ratio of the electron • -1.76 x 10^8 c/g Robert Milliken (1909) • the charge of a single electron • calculated the mass of drops from radio and density • calculated the charge of an electron with his Oil Drop Experiment • • DATA: • -3.20 x 10^-19 C • -6.40 x 10^-19 C • -4.80 x 10^-19 C • FOUND: • The Least Common Multiple of LCM of the data to be -1.60 x 10^-19 C • This is the charge of an electron • *See the in-class example packet. 1 1/11/16 The Subatomic Particles • An atom is made up of subato

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Consider a gas of non interacting spin-0 bosons at high