Concept explainers
Interpretation:
The mean vibrational energy of CCl4 with the wavenumbers, 459 cm−1, 217 cm−1, 776 cm−1, 314 cm−1 and has to be calculated at different temperatures. The graph between mean vibrational energy of CCl4 at different values of temperatures has to be drawn. The temperature at which the equipartition value within 5% is accurate has to be stated.
Concept introduction:
Statistical thermodynamics is used to describe all the possible configurations in a system at given physical quantities such as pressure, temperature, and a number of particles in the system. The important quantity in the thermodynamic is partition function that is given by the expression as shown below.
q≡∑igie−β∈i
The mean vibrational energy per mode is given by the expression as shown below.
〈εV〉=hc˜νehc˜ν/kT−1
Answer to Problem 13C.6AE
The mean vibrational energy of CCl4 is calculated at different temperatures as shown below in the table.
Temperature (K) | The mean vibrational energy of CCl4 (J) |
1000 | 8.97×10−20 |
2000 | 2.11×10−19 |
3000 | 3.35×10−19 |
4000 | 4.59×10−19 |
5000 | 5.83×10−19 |
6000 | 7.07×10−19 |
The graph between mean vibrational energy of CCl4 at different values of temperatures is shown below.
The temperature at which the equipartition value within 5% is accurate is 6000 K.
Explanation of Solution
The mean vibrational energy per mode is given by the expression as shown below.
〈εV〉=hc˜νehc˜ν/kT−1
Where,
- k is the Boltzmann constant(1.38×10−23 J K−1).
- T is the temperature.
- h is the Planck’s constant. (6.62608×10−34 J s)
- c is the speed of light. (2.997945×1010 cm s−1)
- ˜ν is the vibrational wavenumber.
The table that represents the wavenumber and degeneracy of the stated of CCl4 is shown below.
Wavenumber | Degeneracy |
459 cm−1 | 1 |
217 cm−1 | 2 |
776 cm−1 | 3 |
314 cm−1 | 3 |
The mean vibrational energy for CCl4 is given by the expression as shown below.
〈εV〉=hc˜ν1ehc˜ν1/kT−1+(2)hc˜ν2ehc˜ν2/kT−1+(3)hc˜ν3ehc˜ν3/kT−1+(3)hc˜ν4ehc˜ν4/kT−1 (1)
Substitute the values of wavenumber at different levels, k, c, and h in the equation (1).
〈εV〉=((6.62608×10−34 J s)(2.997945×1010 cm s−1)(459 cm−1)e(6.62608×10−34 J s)(2.997945×1010 cm s−1)(459 cm−1)/(1.38×10−23 J K−1)T−1+(2)(6.62608×10−34 J s)(2.997945×1010 cm s−1)(217 cm−1)e(6.62608×10−34 J s)(2.997945×1010 cm s−1)(217 cm−1)/(1.38×10−23 J K−1)T−1+(3)(6.62608×10−34 J s)(2.997945×1010 cm s−1)(776 cm−1)e(6.62608×10−34 J s)(2.997945×1010 cm s−1)(776 cm−1)/(1.38×10−23 J K−1)T−1+(3)(6.62608×10−34 J s)(2.997945×1010 cm s−1)(314 cm−1)e(6.62608×10−34 J s)(2.997945×1010 cm s−1)(314 cm−1)/(1.38×10−23 J K−1)T−1)
The above expression can be simplified as shown below.
〈εV〉=(6.626×10−21e6.626×10−21/(1.38×10−23 J K−1)T−1+(2)4.3106×10−21e4.3106×10−21/(1.38×10−23 J K−1)T−1+(3)1.5415×10−20e1.5415×10−20/(1.38×10−23 J K−1)T−1+(3)6.2375×10−21e6.2375×10−21/(1.38×10−23 J K−1)T−1) J〈εV〉=(6.626×10−21e480.14/T−1+8.6212×10−21e312.36/T−1+4.6245×10−20e1117.03/T−1+18.8025×10−21e451.99/T−1) J
Substitute the value of T=1000 K in the above expression.
〈εV〉=(6.626×10−21e480.14/1000−1+8.6212×10−21e312.36/1000−1+4.6245×10−20e1117.03/1000−1+18.8025×10−21e451.99/1000−1) J=(6.626×10−211.6163−1+8.6212×10−211.3666−1+4.6245×10−203.0558−1+18.8025×10−211.5714−1) J=(1.0751×10−20+2.3517×10−20+2.2495×10−20+3.2906×10−20) J=8.9669×10−20 J
Similarly, the mean vibrational energy of CCl4 is calculated at different temperatures as shown below in the table.
Temperature (K) | The mean vibrational energy of CCl4 (J) |
1000 | 8.97×10−20 |
2000 | 2.11×10−19 |
3000 | 3.35×10−19 |
4000 | 4.59×10−19 |
5000 | 5.83×10−19 |
6000 | 7.07×10−19 |
The graph between mean vibrational energy of CCl4 at different values of temperatures is shown below.
Figure 1
The equipartition value of CCl4 is 9kT.
The equipartition value of CCl4 is calculated at T=1000 K as shown below.
ε=9kT=9(1.38×10−23 J K−1)(1000 K)=1.2429×10−19 J
The equipartition value of CCl4 is calculated at different temperatures as shown below in the table.
Temperature (K) | The equipartition value of CI4 (J) |
1000 | 1.2429×10−19 |
2000 | 2.4858×10−19 |
3000 | 3.7287×10−19 |
4000 | 4.9716×10−19 |
5000 | 6.2145×10−19 |
6000 | 7.46×10−19 |
It is expected that the difference between the equipartition value and the mean vibrational energy of CCl4 at 6000 K is 5%.
The percentage error in the value is calculated as shown below.
ΔE%=|ε1−ε2ε1|×100
Where,
- ε1 is the equipartition value of CCl4.
- ε2 is the mean vibrational energy of CCl4.
Substitute the value of ε1 and ε2 in the above equation.
ΔE%=|7.46×10−19 J−7.07×10−197.46×10−19 J|×100%=0.397.46×100%≈5%
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Chapter 13 Solutions
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