How does the Equipartition Theorem help to evaluate the heat capacity of a cubic meter of ordinary room air?

Question

Don't calculate it here, merely explain what you need to do in order to calculate it. Give the strategy, not the result.

Answer 1

The Equipartition Theorem says that each energy mode (actually, degree of freedom) holds the same amount of energy:

http://en.wikipedia.org/wiki/Equipartition_theorem

Consider a simple, monatomic, molecule such as helium. The only energy mode is the kinetic energy of the molecule as a whole, with three degrees of freedom (X, Y, and Z). Thus the energy of an individual molecule is (3/2)kT.

Compare this with a simple diatomic molecule (simple because it does not support vibration). Now, in addition to the kinetic energy of the molecule as a whole, there is also the rotational kinetic energy of molecule around its center of gravity (two additional degrees of freedom). Thus the energy of an individual molecule is going to be (5/2)kT.

http://www.plmsc.psu.edu/~www/matsc597c-1997/systems/Lecture4/node1.html

Suppose we now add a given amount of energy to N molecules of each, individually. The same energy goes into each molecule, lets call it dE, generates a rise in temperature of dT.

For the first molecule type:

E = (3/2)kT so
E + dE = (3/2)k(T + dT) or
dE = (3/2) k dT

while for the second molecule, using the same logic, dE = (5/2) k dT.

The "heat capacity" is, by definition dE/dT so the heat capacity of the two types of molecules is different.

Ordinary room air is a mixture of gases, each of different type. The temperature is the same for all, but because the gases are different, their specific heats are different:

hydrogen: 28.82
helium and argon: 20.7862
water vapor: 37.47
nitrogen: 29.12
etc.


The specific heat of the total is the weighted sum of the individual specific heats. Since nitrogen and oxygen predominate, you'd expect the result to be very close to that of those two gases, and it it is, but let's be more general.

So, start with the number of molecules of each gas in a cubic meter of ordinary room air. Determine the specific heat of each, either using the Equipartition Theorem or a table of specific heats. Determine the amount of energy needed to raise the molecules of each component by one degree. Take the sum to get the total energy needed to raise the temperature of the air. Then adjust to get the specific quantity you want (specific heat capacity, heat capacity for the full cubic meter, etc.)

Of course there are shortcuts, but there is no point to using them until you understand the basics.

(In practice, you would never use the Equipartition Theorem directly, but use the more accurate, measured values, etc.)

Answer 2

Equipartition Theorem