Question by Student 201427145 Professor, I have thought about Thermally Equilibrium that you had mentioned at last class. We can assume initially Diaphram X(closed system) divided in two Volume by membrane and each has Gas A, B. And then membrane has eliminated and Gas A and B mixed perfectly. Diaphram X, gas A, B have energy of $$E_{X}, E_{A}, E_{B}$$ each other. First, $$Define:\Omega(E)\equiv$$[Number of Microstate according to Energy]. If system has Energy(E), what we can see is macrostate but it is one of a lot of microstate case of probability distribution. Macrostate can be easily occured at E which makes Maximum number of Microstate. $$\therefore\frac{d\Omega(E_{X})}{dE}=0$$ Here Maximum Number of Microstate occurs at Energy of X, A and B. For the convenient, Let's focus on Energy of A. $$\therefore\frac{d\Omega(E_{X})}{dE_{A}}=0$$ Note: Closed system- $$E_{X}=E_{A}+E_{B} \Rightarrow \Omega(E_{X})=\Omega(E_{A}+E_{B})$$ State A and B is independent- $$\Omega(E_{A}+E_{B})=\Omega(E_{A})\Omega(E_{B})\Rightarrow \Omega(E_{X})=\Omega(E_{A})\Omega(E_{B})$$ $$\therefore \frac{d\Omega(E_{x})}{dE_{A}}=\frac{d\Omega(E_{A})\Omega(E_{B})}{dE_{A}}=\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{A}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{B}}\frac{dE_{B}}{dE_{A}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=0$$ $$Note: E_{X}=E_{A}+E_{B}=const\Rightarrow dE_{A}+dE_{B}=0 \Rightarrow\frac{dE_{B}}{dE_{A}}=-1$$ $$\therefore -\Omega(E_{A})\frac{d\Omega(E_{B})}{dE_{B}}+\Omega(E_{B})\frac{d\Omega(E_{A})}{dE_{A}}=0$$ $$\frac{1}{\Omega(E_{A})}\frac{d\Omega(E_{A})}{dE_{A}}=\frac{1}{\Omega(E_{B})}\frac{d\Omega(E_{B})}{dE_{B}}$$ $$Note: \frac{dln{\Omega(E)}}{d\Omega(E)}=\frac{1}{\Omega(E)} \Rightarrow\frac{d\Omega(E)}{\Omega(E)}=dln{\Omega(E)}$$ $$\therefore\frac{dln{\Omega(E_{A})}}{d\Omega(E_{A})}=\frac{dln{\Omega(E_{B})}}{d\Omega(E_{B})}$$ $$Define: \frac{1}{k_{B}T}\equiv\frac{dln{\Omega(E)}}{d\Omega(E)}$$ where $$k_{B}$$ is Boltzmann constant $$\therefore\frac{dln{\Omega(E_{A})}}{d\Omega(E_{A})}=\frac{dln{\Omega(E_{B})}}{d\Omega(E_{B})}=\frac{1}{k_{B}T}$$ $$\therefore T_{A}=T_{B}=T-Answer$$
This is a good effort. It's a bit unclear whether $$\therefore\frac{d\Omega(E_{X})}{dE_{A}}=0$$ is correct. But it's on the right track.
 $\pi$