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Q. 6.51

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An Introduction to Thermal Physics
Found in: Page 256
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

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 1h3 to 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.

The translational partition function is given by Ztr=VvQ.

See the step by step solution

Step by Step Solution

Step 1. Given information

The translational partition function is given by

Ztr=1h3d3rd3pe-EtrkT...........................(1)

Step 2. Calculation

As the translational kinetic energy of the molecules does not depend on the position, the integrand is also independent of the position and the integral over the position variables only yields the total volume V of the box.

The expression for the translational energy of the gas molecules is given by

Etr=px22m+py22m+pz22m..................(2)

Here, x-,y- and z- denotes the three components of the momentum.

Substitute the expression of energy into equation (1) and solve to calculate the value of the translational partition function.

role="math" localid="1647147921072" Ztr=Vh30dpxe-px22mkT0dpye-py22mkT0dpze-pz22mkT=Vh3π·2mkTπ·2mkTπ·2mkT=Vh32πmkT32........................(3)

The quantum volume of the gas molecules is given by

vQ=h22πmkT32........................(4)

Substitute the value of the quantum volume from equation (4) into equation (3) to obtain the required translational partition function.

Ztr=Vh32πmkT32=V1h22πmkT32=VvQ

Most popular questions for Physics Textbooks

Consider a classical particle moving in a one-dimensional potential well ux, as shown. The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

x=xe-βuxdxe-βuxdx

where each integral is over the entire xaxis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(z) in a Taylor series about the equilibrium point

ux=ux0+x-x0dudxx0+12x-x02d2udx2x0+13!x-x03d3udx3x0+........

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x=x0.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for x become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series ux

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

ux=u0x0x12-2x0x6

Sketch this function, and show that the minimum of the potential well is at x=x0, with depth u0. For argon, x0=3.9A and u0=0.010eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u0. Evaluate the result numerically for argon, and compare to the measured value α=0.0007K-1

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