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These notes and slides are part of a course on thermodynamics.

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- Zhigang Suo's blog
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These notes and slides are part of a course on thermodynamics.

»

- Zhigang Suo's blog
- Log in or register to post comments
- 12075 reads

## Comments

## Question on ideal gas

Zhigang,

For an isolated system consists of ideal gas, I have one question concerning the relation between the total number of states and the volume. For a container of volume V contains N molecules, in the note we argue sigma(U,V) scales to V^N. Now consider two configurations of the system.

Configuration 1: each single molecule is allowed to occupy any position within the container.

Configuration 2: each single molecule is allowed to occupy any position within the container, except a given CUBIC portion of the container of volume dV <<V. The allowed region thus has a volume of V’ =V-dV that is almost equal but smaller than V. Note there is no physical separator to stop molecules from going inside the “vacuum” cube though.

On one side, we know the system is much probable to stay in configuration-1 instead of configuration-2 when in equilibrium due to diffusion. On the other side, since V is almost equal to V’, according to the argument at the beginning, the probability to stay in each configuration should be almost equal. How to reconcile this apparent contradiction?

Let’s consider one more configuration (3): each single molecule is allowed to occupy any position within the container, except a given “weird shaped” portion of the container of volume dV same as in configuration 2. Is it true that the system is equally probable to stay in configuration-2 and configuration-3? From the scaling argument the answer seems to be yes, but it is quite anti-intuitive, as least to me.

Thank you very much.

Li Han

## Re: Question on ideal gas

The ratio of the probability for the gas to be in configuration (2) to the probability of the gas in configuration (1) equals (V-dV)^N/V^N. See the example on ink particles.

## Possible explanation from continuum view

Dear Han

I want to share some of my ideas, though I'm not sure if it is right or not. Any way, I hope this would make any help.

I think the description like sigma(U,V) scales to V^N is from the continuum view, though it is not really necessary for the statistics mechanics. It means that there are elements which are infinitesimal in the macroscopic view while are infinite in the microscopic view. So the volume of molecules can be ignored compared that of the elements, denoted v_e. If this is the case, V^N can be explained in the following way. There are (V/v_e) choices for molecules and one single portion v_e can contain all the N molecules. In fact, v_e would be concealed in the finial expression of the probability of certain configuration. Under this understanding, your second question can be answered that no matter what the portion shape is, the choices they have are same, so long as their volumes are equal.

As for the first question, I think that the most probable configuration is the concept of statistics. In the text book

Entropy, Order Parameters, and ComplexityJames P. Sethna, Prof. James P. Sethna presents the distribution of 2N particles in a box as an example to illustrate that almost exactly half of the coordinates are on the right half of their range in almost all the volume of a box in R^6N. (see page 40 for reference). I believe the analysis in the text can be applied to your question in the similar way.(The book can be download through Prof. James P. Sethna's personal web http://www.lassp.cornell.edu/sethna/ directly)

Best Regards!

Teng Zhang

## Thanks

Thanks Zhigang and Teng for the reply.

I just realized this is just a combinatory problem indeed: for the same set of particles, when the volume of the container increases to say 10 times the original size, each particle has 10 times more choices to get located. And because particles are independent, N particles thus have 10^N more choices(states). I think this is what Teng has explained, rephrased in my own words.

Li Han