Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
Question
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Chapter 17, Problem 17.58E
Interpretation Introduction

Interpretation:

The values of partition function with the given temperatures are to be calculated. The validation of the fact that leveling off of the value of q as the temperature increases and the interpretation of the values of q are to be stated.

Concept introduction:

Statistical thermodynamics is used to describe all possible configurations in a system at given physical quantities such as pressure, temperature and number of particles in the system. An important quantity in thermodynamics is partition function that is represented as,

qigiei/kT

Where,

gi represents the degeneracy.

i represents the energy of ith microstate.

k represents the Boltzmann constant with value 1.38×1023J/K.

T represents the temperature (K).

It is also called as canonical ensemble partition function.

Expert Solution & Answer
Check Mark

Answer to Problem 17.58E

The values of partition function at given temperatures 50K, 100K, 200K, 300K, 500K and 1000K is 2.530, 3.431, 4.898, 5.894, 7.080 and 8.320 respectively. The fact that leveling off of the value of q as the temperature increases has been validated.

Explanation of Solution

The given values of temperatures are 50K, 100K, 200K, 300K, 500K and 1000K.

It is given that a system has energy levels at 0, 1×1021J, 2.5×1021J, 4×1021J and 6×1021J.

The degeneracy of each level is 2.

The important quantity in the thermodynamic is partition function that is represented as,

qigiei/kT …(1)

Where,

gi represents the degeneracy.

i represents the energy of ith microstate.

k represents the Boltzmann constant with value 1.38×1023J/K.

T represents the temperature (K).

The partition function of a system that has five different energy levels with energy 0, 1, 2, 3, 4 and 5 is represented as,

q=ge0/kT+ge1/kT+ge3/kT+ge4/kT+ge5/kT

q=g[e0/kT+e1/kT+e3/kT+e4/kT+e5/kT] …(2)

Substitute the value of energies, degeneracies, k and T=50K in the equation (2).

q=2[exp(0J(50K)(1.38×1023J/K))+exp(1×1021J(50K)(1.38×1023J/K))+exp(2.5×1021J(50K)(1.38×1023J/K))+exp(4×1021J(50K)(1.38×1023J/K))+exp(6×1021J(50K)(1.38×1023J/K))]=2[exp(0)+exp(1.449)+exp(3.623)+exp(5.797)+exp(8.696)]2.530

Therefore, the value of q is 2.530 at 50K.

Substitute the value of energies, degeneracies, k and T=100K in the equation (2).

q=2[exp(0J(100K)(1.38×1023J/K))+exp(1×1021J(100K)(1.38×1023J/K))+exp(2.5×1021J(100K)(1.38×1023J/K))+exp(4×1021J(100K)(1.38×1023J/K))+exp(6×1021J(100K)(1.38×1023J/K))]=2[exp(0)+exp(0.725)+exp(1.812)+exp(2.899)+exp(4.348)]3.431

Therefore, the value of q is 3.431 at 100K.

Substitute the value of energies, degeneracies, k and T=200K in the equation (2).

q=2[exp(0J(200K)(1.38×1023J/K))+exp(1×1021J(200K)(1.38×1023J/K))+exp(2.5×1021J(200K)(1.38×1023J/K))+exp(4×1021J(200K)(1.38×1023J/K))+exp(6×1021J(200K)(1.38×1023J/K))]=2[exp(0)+exp(0.362)+exp(0.906)+exp(1.449)+exp(2.174)]4.898

Therefore, the value of q is 4.898 at 200K.

Substitute the value of energies, degeneracies, k and T=300K in the equation (2).

q=2[exp(0J(300K)(1.38×1023J/K))+exp(1×1021J(300K)(1.38×1023J/K))+exp(2.5×1021J(300K)(1.38×1023J/K))+exp(4×1021J(300K)(1.38×1023J/K))+exp(6×1021J(300K)(1.38×1023J/K))]=2[exp(0)+exp(0.242)+exp(0.604)+exp(0.966)+exp(1.449)]5.894

Therefore, the value of q is 5.894 at 300K.

Substitute the value of energies, degeneracies, k and T=500K in the equation (2).

q=2[exp(0J(500K)(1.38×1023J/K))+exp(1×1021J(500K)(1.38×1023J/K))+exp(2.5×1021J(500K)(1.38×1023J/K))+exp(4×1021J(500K)(1.38×1023J/K))+exp(6×1021J(500K)(1.38×1023J/K))]=2[exp(0)+exp(0.145)+exp(0.362)+exp(0.580)+exp(0.870)]7.080

Therefore, the value of q is 7.080 at 500K.

Substitute the value of energies, degeneracies, k and T=1000K in the equation (2).

q=2[exp(0J(1000K)(1.38×1023J/K))+exp(1×1021J(1000K)(1.38×1023J/K))+exp(2.5×1021J(1000K)(1.38×1023J/K))+exp(4×1021J(1000K)(1.38×1023J/K))+exp(6×1021J(1000K)(1.38×1023J/K))]=2[exp(0)+exp(0.0725)+exp(0.1812)+exp(0.2898)+exp(0.4348)]8.320

Therefore, the value of q is 8.320 at 1000K.

On comparing the partition function values it is observed that, the value of partition function increases with the increase in temperature. At low temperatures the increase of 50K results in the increase in partition function of about 1.00 unit, while at higher temperatures with the increase of 500K in temperature, the increase in partition function is of about 1.00 unit. Thus, this shows as the temperature increases the increment in the values of partition function decreases.

Conclusion

The values of partition function at given temperatures 50K, 100K, 200K, 300K, 500K and 1000K is 2.530, 3.431, 4.898, 5.894, 7.080 and 8.320 respectively. The fact that leveling off of the value of q as the temperature increases has been validated.

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Chapter 17 Solutions

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