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Q. 6.48

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An Introduction to Thermal Physics
Found in: Page 255
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

For a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy Ze of the electronic ground state.

(a) Show that the entropy in this case is

S=NkInVZeZrotNvQ+72.

Calculate the entropy of a mole of oxygen at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book.

(b) Calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature. Express the answer in electron-volts

(a) The statement is proved below and the entropy of a mole of oxygen at room temperature and atmospheric pressure isS=205.1J/k.

(b) The chemical potential of oxygen in earth's atmosphere near sea level, at room temperature isμ=-0.547 eV.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Part (a) step 1: Given information

We have given that a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy Ze

We need to calculate the entropy of a mole of oxygen at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book.

Part (a) step 2: Simplify

The entropy is given as:

S=Nk In VNvQ+52-ϑFintϑT

where Fint is the internal Helmholtz free energy and it's given in terms of the internal partition as:

Fint=-Nkt In(Zint)

for a diatomic gas, the internal partition function is simply the rotational partition function multiplied by the degeneracy of the electronic ground state Ze ,the at is:

Zint=ZeZrot

substitute into (2) to get:

Fint=-NkT In(ZeZrot)

the partial derivative with respect to the temperature is, So(note that Zrot is proportional to T ):

ϑFintϑt=-NkIn(ZeZrot0-nk

substitute into (1) to get:

S=Nk InVNvQ+52+nKIn(ZeZrot)+NkS=Nk In VNvq+In(ZeZrot)+52S+NK In VZeZrotNvQ+72 (3)

Considering an oxygen molecule at the room temperature and atmospheric pressure where Ze=3 . The rotation partition for a diatomic gas is given by:

Zrot=kT2∈

Here,∈ is the energy constant, which equals 0.00018 eV for oxygen. So at the room temperature, the rotational partition function is:

The quantum volume will be given as:

vQ=h22πmkT

The mass of the oxygen molecule O2 is 32u ,hereu=1.66-27kg , substituting (not that h=6.626×10-34J.s and k=1.38×10-23J/K ):

vQ=(6.626×10-34J.s)22π(32×1.66×10-27kg)(1.38×10-23J/K)(300K)=5.66×10-33m3

the volume per particle is written as (from the ideal gas law):

VN=kTP

at a pressure of 1 atm and the room temperature we have:

vN=(1.38×10-23J/K)(300K)1.01×105Pa=4.1×10-26m3

plug all these numbers into the natural logarithm in the equation (3) to get:

InVZeZrotNvQ=In (4.1×10-26m3)(3)(71.8)5.66×10-33m3=21.17

now substitute into the equation (3) to get the entropy as:

S=Nk 21.17+72

but Nk=nR ,

S=n(8.314J/K.mol) 21.17+72

for one mole n=1 ,

S=(8.314J/K) 21.17+72=205.1J/KS=205.1 J/K

Part (b) step 1: Given information

We have given that a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy Ze of the electronic ground state.

We need to calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature.

Part (b) step 2: Simplify

The chemical potential is given by:

μ=-kTInVZeZrotNvQ

substitute with the values to get:

μ=-(8.62×10-5eV/K)(300K)(21.17)=-0.547 eVμ=-0.547 eV

Most popular questions for Physics Textbooks

A particle near earth's surface traveling faster than about 11 km/s has enough kinetic energy to completely escape from the earth, despite earth's gravitational pull. Molecules in the upper atmosphere that are moving faster than this will therefore escape if they do not suffer any collisions on the way out.

(a) The temperature of earth's upper atmosphere is actually quite high, around 1000 K. Calculate the probability of a nitrogen molecule at this temperature moving faster than 11km/s, and comment on the result.

(b) Repeat the calculation for a hydrogen molecule (H2) and for a helium atom, and discuss the implications.

(c) Escape speed from the moon's surface is only about 2.4 km/s. Explain why the moon has no atmosphere.

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 1013 Hz. 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 problem 6.20 you computed the partition function for a quantum harmonic oscillator :Zh.o.=11-e-βε, where ε=hf is the spacing between energy levels.

(a) Find an expression for the Helmholtz free energy of a system of N harmonic oscillators.

(b) Find an expression for the entropy of this system as a function of temperature. (Don't worry, the result is fairly complicated.)

In this problem you will investigate the behavior of ordinary hydrogen, H2, at low temperatures. The constant ε is0.0076eV. As noted in the text, only half of the terms in the rotational partition function, equation 6.3, contribute for any given molecule. More precisely, the set of allowed j values 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 Problem 6.28) 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 j are 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 1/4 parahydrogen and 3/4 orthohydrogen. 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/2 per 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 D2 gas, consisting of a normal equilibrium mixture of 2/3 ortho 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.*

At room temperature, what fraction of the nitrogen molecules in the air are moving at less than 300 m/s?
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