2
$\begingroup$

The formula for the multiplicity of a monoatomic ideal gas is

$$\qquad \Omega(U,V,N)=f(N)V^NU^{3N/2}$$

where $V$ is the volume of position space, $U$ is the molecule's kinetic energy, $N$ is the number of indistinguishable gas molecules and $f(N)$ is a complicated function of N

And from Schroeder's An Introduction to Thermal Physics:

Sometimes you can calculate probabilities of various arrangements of molecules just by looking at the volume dependence of the multiplicity function. For instance, suppose we want to know the probability of finding the configuration where all the molecules in a container of gas are somewhere in the left half. This arrangement is just a macrostate with the same energy and number of molecules, but half the original volume. Looking at equation we see that replacing $V$ by $V/2$ reduces the multiplicity by a factor of $2^N$. In other words, out of all the allowed microstates, only one in $2^N$ has all the molecules in the left half. Thus the probability of this arrangement is $2^{-N}$

Why do we just simply take $2^{-N}$ as our probability? Shouldn't we consider the macrostate for other volumes as well and make the probability be

$$Pr=\frac{\Omega(U,V/2,N)}{\sum_{V=0}^{100} \Omega(U,V,N)} \ ?$$

$\endgroup$
1
$\begingroup$

Not really.

You want to calculate the probability to find all the molecules in the left half of the volume.

Since the probability density is

$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$

where, $c_N$ is a constant and, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$

what you need to consider is

$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$

where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.

You can immediately see that

$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$

So that you finally obtain the desired result.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.