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$\int \frac{Q_{rev}}{T} = \Delta(k_B\ln\Omega)=\Delta S$
Could anyone give some definite proof for this?

I was able to prove that the two definitions of change in entropy are equivalent for an isothermal process carried out on a gas (by quantizing space and then limiting the quantization to infinity), but my proof makes the absolute entropy of the gas infinite. If the process is not isothermal, the particle's velocities come into the picture and I don't know how to deal with that. I tried making various assumptions (quantizing time, etc), but it didn't work. I know that once I prove it for another process, it will be proven for any process carried out on ideal gases(as I can write any process as the combination of isothermal and another process).

Could someone please nudge me in the right direction/give a proof?

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In what framework do you work? If you know define the temperature as $\frac{1}{T}=\frac{\partial S}{\partial U}$, then you can just solve the first law of thermodynamics $\delta Q=PV+dU(S,V)$ for $dS$. – Nick Kidman Jan 31 '12 at 14:35
@NickKidman I'm defining everything on the basis of (P,V,T). Then I define $\Delta S$ as $\int\frac{Q}{T}$, and try to prove that this is equal to change in %k_Bln\Omega$ . How does your approach merge these two definitions of entropy? It seems that you are only proving a general formula for entropy (I'm not sure, though) – Manishearth♦ Feb 3 '12 at 2:37
In classical thermodynamics, there are only macroscopical quantities. For every $(P,V,T)$, there are many many microstates. The introduction of microstates only comes with statistical mechanics. In that framework the volume $\Omega(E,V)$ is a natural object in phase space and $S(E,V)=log(\Omega(E,V))$ is the definition of the entropy. You can now compute the derivative $dS$ (where $dE$, i.e. $dU$, is related to $\delta Q$ via first law of theremodynamics, if you will). Both $T$ and $P$ are abstractly defined in terms of derivatives, although $P$ is easy to motivate. – Nick Kidman Feb 3 '12 at 8:17
Yeah, but I want to calculate $\Omega$ as a function of P/V/T/U/whatever. Omega as a func of V is easy at a constant temperature, but when the temperature changes, the speegs of the molecules add to the microstates in a way which I don't know how to calculate. – Manishearth♦ Feb 3 '12 at 10:36
I don't understand. So you're doing calculations within the framework of statistical mechanics after all? If you have a potential like $U$ or $S$, then you can compute the equation of state, and if you like you can express $V$ as function of, for example $T$. – Nick Kidman Feb 3 '12 at 12:26
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1 Answer

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The fact that $k_B \ln \Omega$ coincides with entropy $S$ defined in thermodynamics comes from microcanonical ensemble.

There are many resources out there on microcanonical ensemble, for example, this. After you come to the conclusion that

$$\beta=\left( \frac{ \partial \ln \Omega }{ \partial U }\right)_{N,V}$$

fully characterizes thermal equilibrium, you know that it must be a function of thermodynamic temperature, and thermodynamic temperature alone, by virtue of zeroth law of thermodynamics. So $\beta=f(T)$.

Compare this to

$$\frac{1}{T}=\left( \frac{ \partial S }{ \partial U }\right)_{N,V}$$

and you get that S must be the function of $\ln \Omega$. The remaining question is the exact form of this function, and you already derive it from special cases.

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Could you please provide a more detailed explanation? I'm not too familiar with the statistical half of thermodynamics. I want to try to prove it by counting microstates in a system given P,V,T. If this is not possible, do you have a link/etc to the proof? – Manishearth♦ Feb 3 '12 at 2:42
By the way, S as a function of $\ln\Omega$ can be more easily proved from the multiplicativeness of probability and from the fact that S is extensive. Not exactly the type of answer I was looking for, but it'll do. Nick Kidman gave a nice explanation above that clarified my main confusion. – Manishearth♦ Feb 6 '12 at 4:10
@Manishearth: How do you prove S as a function of $\ln \Omega$ from the multiplicativeness of probability and from the fact that S is extensive? – C.R. Feb 6 '12 at 9:47
If I mix two systems with $\Omega_1$, $\Omega_2$, the resultant $\Omega=\Omega_1\times\Omega_2$. But, Entropy is extensive, so $S=S_1+S_2$. If $S_r=f(\Omega_r)$, then $f(x)=k\ln x$ (only function which turns multiplication into addition), thus $S=k\ln\Omega$. We can later set the constant $k=k_B$. – Manishearth♦ Feb 6 '12 at 12:05
@Manishearth: But you need to prove that S is indeed related to Ω, and not to any other quantities. U is also extensive, but you don't get $U=k_B \ln \Omega$. – C.R. Feb 6 '12 at 12:40
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