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Q 6.10

Expert-verifiedFound in: Page 228

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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 10^{13 }Hz. As
for any quantum harmonic oscillator, the energy levels are $\frac{1}{2}hf,\frac{3}{2}hf,\frac{5}{2}hf$, 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).

The probability of water molecule is:

${P}_{1}=0.9997\phantom{\rule{0ex}{0ex}}{P}_{2}=4.618\times {10}^{-4}\phantom{\rule{0ex}{0ex}}{P}_{3}=2.133\times {10}^{-7}$

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 10^{13 }Hz. As for any quantum harmonic oscillator, the energy levels are $\frac{1}{2}hf,\frac{3}{2}hf,\frac{5}{2}hf$, and so on. None of these levels are degenerate.

(a) The partition function is: because none of the levels are degenerate.

$Z=\sum _{s}{e}^{-E\left(s\right)/kT}$

Where,

$E\left(s\right)=\left(s+\frac{1}{2}\right)hfs=0,1,2,\dots $

So,

$Z={e}^{-hf/2kT}+{e}^{-3hf/2kT}+{e}^{-5hf/2kT}+\dots $

Let $x=hf/kT$, so

$Z={e}^{-x/2}+{e}^{-3x/2}+{e}^{-5x/2}+\dots \left(1\right)$

As a result, the value of z is (at T = 300 K, we substitute the Boltzmann constant in eV, k = 8.617x 10^{-5} eV/K, and the Planck constant in eV, h = 4.136 x 10^{-15} eV s):

$x=\frac{\left(4.136\times {10}^{-15}\mathrm{eV}\xb7\mathrm{s}\right)\left(4.8\times {10}^{13}\mathrm{Hz}\right)}{\left(8.617\times {10}^{-5}\mathrm{eV}\right)(300\mathrm{K})}=7.68$

Substitute x into (1)

$Z={e}^{-(7.68)/2}+{e}^{-3(7.68)/2}+{e}^{-5(7.68)/2}\phantom{\rule{0ex}{0ex}}Z=0.0215$

Probability of first state is:

$P=\frac{1}{Z}{e}^{-x/2}$

Substitute with x and z:

role="math" localid="1647366550518" ${P}_{1}=\left(\frac{1}{0.0215}\right){e}^{-7.68/2}\phantom{\rule{0ex}{0ex}}{P}_{1}=0.9997\phantom{\rule{0ex}{0ex}}$

Probability of second state is:

${P}_{2}=\frac{1}{Z}{e}^{-3x/2}$

Substitute x and z:

role="math" localid="1647366573852" ${P}_{2}=\left(\frac{1}{0.0215}\right){e}^{-3(7.68)/2}\phantom{\rule{0ex}{0ex}}{P}_{2}=4.618\times {10}^{-4}$

Probability of third state:

${P}_{3}=\frac{1}{Z}{e}^{-3x/2}$

Substitute x and z:

role="math" localid="1647366591263" ${P}_{3}=\left(\frac{1}{0.0215}\right){e}^{-5(7.68)/2}\phantom{\rule{0ex}{0ex}}{P}_{3}=2.133\times {10}^{-7}$

We have temperature value, T=300k, so

$x=\frac{\left(4.136\times {10}^{-15}\mathrm{eV}\xb7\mathrm{s}\right)\left(4.8\times {10}^{13}\mathrm{Hz}\right)}{\left(8.617\times {10}^{-5}\mathrm{eV}\right)(700\mathrm{K})}=3.2913$

Substitute x into (1):

role="math" localid="1647366281815" $Z={e}^{-(3.2913)/2}+{e}^{-3(3.2913)/2}+{e}^{-5(3.2913)/2}\phantom{\rule{0ex}{0ex}}Z=0.20033\phantom{\rule{0ex}{0ex}}$

Substitute the value of x and z to get the probabilities:

${P}_{1}=\left(\frac{1}{0.20033}\right){e}^{-3.2913/2}\phantom{\rule{0ex}{0ex}}{P}_{1}=0.962847\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{P}_{2}=\left(\frac{1}{0.20033}\right){e}^{-3(3.2913)/2}\phantom{\rule{0ex}{0ex}}{P}_{2}=0.035823\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{P}_{3}=\left(\frac{1}{0.20033}\right){e}^{-5(3.2913)/2}\phantom{\rule{0ex}{0ex}}{P}_{3}=0.001333\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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