GREPhysics.NET: GR9677 #79

GR9677 #79

Problem

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Statistical Mechanics $\Rightarrow$ }Specific Heat

The specific heat at constant volume for high temperatures is $c_v = 7/2 R$ . The specific heat at low temperatures is $3/2 R$ . Why?

There are three contributions to the specific heat of a diatomic gas. There is the translational, vibrational, and rotational. At low temperatures, only the translational heat capacity contributes $U = 3/2 N k T \approx c_v T \Rightarrow c_V=3/2 Nk$ . At high temperatures, all three components contribute, and one has $c_V=(3/2+1+1)Nk = 7/2 Nk$ .

The general formula is $c_v = c_v(translational)+c_v(rotational)+c_v(vibrational)=Nk\left(3/2 + 1 + (h\nu/(k\theta))^2 \exp(h\nu/(k\theta))/(\exp(h\nu/(k\theta))-1)^2 \right)$

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Comments

bluejay27
2016-12-25 22:22:27

Check page 606 and 607. D is the correct answer.

bluejay27
2016-12-25 22:22:53

of Young and Freedman

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hellothar
2012-10-21 20:26:23

No one seems to have pointed out that Cv, for a DIatomic gas, at low temperatures is not 3/2R but 5/2R. (Check HRK, or Fermi, or anywhere else).

Yosun's 3/2R applies only to monoatomic molecules. The problem asks about diatomic molecules.

hellothar
2012-10-21 20:33:06

Nope. Nevermind. The standard texts are assuming only 3+2 vibrational modes. For very, very low temperatures, 3/2 is right.

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shafatmubin
2009-11-04 15:17:49

This link might help:

http://hyperphysics.phy-astr.gsu.edu/HBASE/hframe.html

It shows the variation of Cv of hydrogen gas with temperature. At very high temperatures, above 1000 K, there are 7 degrees of freedom because two additional degrees (apart from the 3 translational and 2 rotational degrees) become active. Can anyone explain what are these two new degrees of freedom? One of them is possibly vibrational, but what about the other?

hdcase
2009-11-04 21:46:12

Hyperphysics doesn't make this very clear, but wikipedia does:
http://en.wikipedia.org/wiki/Specific_heat_capacity#Diatomic_gas

The vibrational state (there is one degree of freedom from it, for a total of 6 degrees of freedom) adds R (instead of R/2) because it carries both a potential and kinetic term (both of which can "hold energy," which is essentially what the specific heat measures).

shafatmubin
2009-11-05 13:12:22

Thanks hdcase. Vibrational energy contributing R instead of 2R makes things a lot more understandable.

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petr1243
2008-03-18 11:45:15

Roughly, Rotaional effects disappear below 25k, and all 3 effects will produce a specific heat of $\frac{7}{2}\tau$ at temperatures above 8000K. A good reference is ch.18 of the University Physics text.

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cyberdeathreaper
2007-02-03 15:05:37

I found this equation helpful in solving the problem:

$C_{V,m} = \frac{1}{2} (3 + v_R + 2v_V)R$

Where

$v_R = 2$

for linear molecules, and

$v_R = 3$

for non-linear molecules.

I assumed that that

$v_V = 1$

for very high temperatures. And of course, the rotational and vibrational contributions are 0 at very low temperatures, as Yosunism explained in the answer.

faith
2010-11-11 20:24:18

equipartition says each degree of freedom has 0.5 K.E.

translational has 3 mode of d.f

rotational
for diatomic ( linear) 2 mode
triatomic has 3x2 mode =6
polyatomic 3 mode (why ? i do not know)

vibrational (i need help completing this)
diatomic = 1 (or 2) mode?
triatomic=?
poly=?

to the question
finding the df for high tem
1/2(3+2+1?)=1?
if its vibrational has 2 mode: 1/2(3+2+2)=7/2

so the answer could be D or E. any formula for vibrational mode under high temp?

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Post A Comment!

You are replying to:

I found this equation helpful in solving the problem:
$C_{V,m} = \frac{1}{2} (3 + v_R + 2v_V)R$
Where
$v_R = 2$
for linear molecules, and
$v_R = 3$
for non-linear molecules.
I assumed that that
$v_V = 1$
for very high temperatures. And of course, the rotational and vibrational contributions are 0 at very low temperatures, as Yosunism explained in the answer.

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