Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

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 $\frac{1}{2} h f, \frac{3}{2} h f, \frac{5}{2} h f$ , 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:

$P_{1} = 0.9997 P_{2} = 4.618 \times 10^{- 4} P_{3} = 2.133 \times 10^{- 7}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

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 $\frac{1}{2} h f, \frac{3}{2} h f, \frac{5}{2} h f$ , and so on. None of these levels are degenerate.

Step 2: Explanation

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

$Z = \sum_{s} e^{- E (s) / k T}$

Where,

$E (s) = (s + \frac{1}{2}) h f s = 0, 1, 2, \dots$

So,

$Z = e^{- h f / 2 k T} + e^{- 3 h f / 2 k T} + e^{- 5 h f / 2 k T} + \dots$

Let $x = h f / k T$ , so

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

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):

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

Substitute x into (1)

$Z = e^{- (7.68) / 2} + e^{- 3 (7.68) / 2} + e^{- 5 (7.68) / 2} Z = 0.0215$

Probability of first state is:

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

Substitute with x and z:

role="math" localid="1647366550518" $P_{1} = (\frac{1}{0.0215}) e^{- 7.68 / 2} P_{1} = 0.9997$

Probability of second state is:

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

Substitute x and z:

role="math" localid="1647366573852" $P_{2} = (\frac{1}{0.0215}) e^{- 3 (7.68) / 2} P_{2} = 4.618 \times 10^{- 4}$

Probability of third state:

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

Substitute x and z:

role="math" localid="1647366591263" $P_{3} = (\frac{1}{0.0215}) e^{- 5 (7.68) / 2} P_{3} = 2.133 \times 10^{- 7}$

Step 3: Explanation

We have temperature value, T=300k, so

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

Substitute x into (1):

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

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

$P_{1} = (\frac{1}{0.20033}) e^{- 3.2913 / 2} P_{1} = 0.962847 P_{2} = (\frac{1}{0.20033}) e^{- 3 (3.2913) / 2} P_{2} = 0.035823 P_{3} = (\frac{1}{0.20033}) e^{- 5 (3.2913) / 2} P_{3} = 0.001333$

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