# Homework Help: Two thermodynamics conceptual questions

1. Oct 18, 2006

### quasar987

There are two conceptual questions from last year's exam that I don't know the answer to.

1) From the point of vue of statistical mechanics, why can't we completely neglect the interaction btw particles, even in an ideal gas?

My best shot is "Though we can neglect the potential energy btw the particles, we still have to account for the collisions btw them." But I don't recall of a place in what we have covered so far about statistical mechanics, where we did account for the collisions btw the particles of a system.

2) Why is the specific heat of a polyatomic gas greater then that of a monoatomic one?

This information is contained within the text, but upon further inspection, I don't see how the explanation given explains the phenomenon. The explanation given in the text is that the polyatomic molecules have rotational degrees of freedom additionally to their translation degree of freedom, so a part of the energy goes into rotational energy, which considerably augment the number of accessible states.

But the specific heat is given by $C=dQ/dT$. Alright, so say a quantity dQ of heat is transfered to the polyatomic gaz. Then its energy goes up by dQ. The temperature of the gaz is defined by

$$kT\equiv \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}$$

(where $\Omega$ is the number of accesible states). We also know that for an ideal gaz, T is function of E only, so

$$kdT = \frac{dT}{dE}dE = \frac{d}{dE}\left( \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}\right) dE = -\frac{1}{\left( \frac{\partial \ln(\Omega)}{\partial E} \right)^2} \frac{\partial^2 \ln(\Omega)}{\partial E^2}dE$$

Granted, $$\frac{\partial \ln(\Omega)}{\partial E}$$ is greater for a polyatomic gaz than for a monoatomic one, but I know nothing about the second derivative of $\ln(\Omega)$, other than it must be negative for dT to be positive as it should. So, is there something I don't see?

Btw - I would appreciate an answer before 12 hours from now, because the exam is in 12h30 and I live 15 minutes away. :-)

Last edited: Oct 18, 2006
2. Oct 19, 2006

### quinn

1) if you ignored interaction between particles then you would never redistrubute enrgy in the gas, and thus you could have two gases mix and they would never thermalize, that is to say both reach the same temperature

2) C = dQ/dT ; therefore C increases when more energy is required to increase temperature. if there are more places/resevoirs for the energy one inputs to go then it will require more energy to increase temperature. The equi-partition therom states that no one state has a preference for energy storage, that is all states must on a statistical average have the same average energy in equilibrium.

To restate more states are accessible to gas(one) than gas(two), gas(one) has more enrgy to add before the temperature goes up because all states can carry the same amount of energy (think: each state is the same size),... -hence discontinuity in C at phase transitions (where new states are suddenly accessible)

Last edited: Oct 19, 2006
3. Oct 19, 2006

### ppithermo

The partition fonction for a polyatomic gas has additional terms. A monoatomic gas only has translation. Polyatomic gases have in addition vibration and rotation.

4. Oct 19, 2006

### Andrew Mason

1) The Maxwell Boltzmann distribution and Ideal Gas law assume elastic collisions. If the molecules have some interaction that uses energy, the collisions are not all perfectly elastic.

2) Temperature is proportional to molecular translational speed^2. Rotational energy does not contribute to temperature.

AM

5. Oct 19, 2006

### StatMechGuy

(1) Another thing is that the equations of state for a classical ideal gas are "poorly behaved". I recall having to do a problem in which we looked at the isothermal compressibility or some such quantity (I think I'm forgetting which one) and the result was that it was divergent. It goes back to the fact that $$U = \frac{3}{2} N k_B T$$. There's no volume term! In fact, it's not immediately obvious how to get the volume term to appear in a quantum ideal gas, but as a hint it involves looking to where you went from an infinite series to an infinite integral, and attempting to evaluate the area you leave behind. If you do this, you do get a volume dependence.

(2) Andrew Mason seems to have hit on this point. Temperature is a measure of translational kinetic energy, so having rotational degrees of freedom sucks up energy that would otherwise go towards increasing the temperature.