Consider a large system of Nindistinguishable, noninteracting molecules (perhaps an ideal gas or a dilute solution). Find an expression for the Helmholtz free energy of this system, in terms of Z1, the partition function for a single molecule. (Use Stirling's approximation to eliminate the N!.) Then use your result to find the chemical potential, again in terms ofZ1.

Short Answer

Expert verified

F=-NkTlnZ1N+1

μ=-kTlnZ1N

Step by step solution

01

Step1. Given information

The partition functionZ for a system of Nindistinguishable, noninteracting molecules is given by

role="math" localid="1647058886923" Z=1N!Z1N...........................(1)

02

Step 2. Calculation for Helmholtz free energy

The formula to calculate the Helmholtz free energy of the system is given by

F=-kTlnZ.............(2)

Substitute the value for Zfrom equation (1) into equation (2) and simplify to obtain the free energy.

role="math" localid="1647059254929" F=-kTln1N!Z1N=-kTNlnZ1-lnZ!=-kTNlnZ1-NlnN+N=-NkTlnZ1N+1...................(3)

03

Step 3. Calculation for chemical potential

The formula to calculate the chemical potential of the system is given by

μ=FNT,V......................(4)

Substitute the value of the free energy from equation (3) into equation (4) and simplify to obtain the chemical potential of the system.

role="math" localid="1647059391759" μ=-NkTlnZ1N+1N=-kTlnZ1N+1-NkTN-lnN=-kTlnZ1N

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Most popular questions from this chapter

The energy required to ionise a hydrogen atom is 13.6 eV, so you might expect that the number of ionised hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionised hydrogen is much larger, nearly one atom in 10,000. Explain why this result is not a contradiction, and why it would be incorrect to try to calculate the fraction of ionised hydrogen using the methods of this section.

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing' mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 1013Hz. As for any quantum harmonic oscillator, the energy levels are 12hf,32hf,52hf, and so on. None of these levels are degenerate.

(a) state and in each of the first two excited states, assuming that it is in equilibrium with a reservoir (say the atmosphere) at 300 K. (Hint: Calculate 2 by adding up the first few Boltzmann factors, until the rest are negligible.) Calculate the probability of a water molecule being in its flexing ground

(b) Repeat the calculation for a water molecule in equilibrium with a reservoir at 700 K (perhaps in a steam turbine).

In this problem you will investigate the behavior of ordinary hydrogen, H2, at low temperatures. The constant εis0.0076eV. As noted in the text, only half of the terms in the rotational partition function, equation6.3, contribute for any given molecule. More precisely, the set of allowed jvalues is determined by the spin configuration of the two atomic nuclei. There are four independent spin configurations, classified as a single "singlet" state and three "triplet" states. The time required for a molecule to convert between the singlet and triplet configurations is ordinarily quite long, so the properties of the two types of molecules can be studied independently. The singlet molecules are known as parahydrogen while the triplet molecules are known as orthohydrogen.

(a) For parahydrogen, only the rotational states with even values of j are allowed.Use a computer (as in Problem6.28) to calculate the rotational partition function, average energy, and heat capacity of a parahydrogen molecule. Plot the heat capacity as a function of kT/t

(b) For orthohydrogen, only the rotational states with odd values of jare allowed. Repeat part (a) for orthohydrogen.

(c) At high temperature, where the number of accessible even-j states is essentially the same as the number of accessible odd-j states, a sample of hydrogen gas will ordinarily consist of a mixture of 1/4parahydrogen and 3/4orthohydrogen. A mixture with these proportions is called normal hydrogen. Suppose that normal hydrogen is cooled to low temperature without allowing the spin configurations of the molecules to change. Plot the rotational heat capacity of this mixture as a function of temperature. At what temperature does the rotational heat capacity fall to half its hightemperature value (i.e., to k/2per molecule)?

(d) Suppose now that some hydrogen is cooled in the presence of a catalyst that allows the nuclear spins to frequently change alignment. In this case all terms in the original partition function are allowed, but the odd-j terms should be counted three times each because of the nuclear spin degeneracy. Calculate the rotational partition function, average energy, and heat capacity of this system, and plot the heat capacity as a function of kT/t.

(e) A deuterium molecule, D2, has nine independent nuclear spin configurations, of which six are "symmetric" and three are "antisymmetric." The rule for nomenclature is that the variety with more independent states gets called "ortho-," while the other gets called "para-." For orthodeuterium only even-j rotational states are allowed, while for paradeuterium only oddj states are allowed. Suppose, then, that a sample of D2gas, consisting of a normal equilibrium mixture of 2/3ortho and 1/3para, is cooled without allowing the nuclear spin configurations to change. Calculate and plot the rotational heat capacity of this system as a function of temperature.*

In this section we computed the single-particle translational partition function,Ztr, by summing over all definite-energy wave functions. An alternative approach, however, is to sum over all position and momentum vectors, as we did in Section 2.5. Because position and momentum are continuous variables, the sums are really integrals, and we need to slip a factor of 1h3to get a unitless number that actually counts the independent wavefunctions. Thus we might guess the formula

role="math" localid="1647147005946" Ztr=1h3d3rd3pe-EtrkT

where the single integral sign actually represents six integrals, three over the position components and three over the momentum components. The region of integration includes all momentum vectors, but only those position vectors that lie within a box of volume V. By evaluating the integrals explicitly, show that this expression yields the same result for the translational partition function as that obtained in the text.

Show explicitly from the results of this section thatG=Nμfor an ideal gas.

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