Equilibrium between a system and a heat reservoir

Question

In Pathria's Statistical Mechanics, Section 3.1, the expression for the probability, $P_r$ of finding a system characterized by the energy value $E_R$ in a reservoir is derived. The derivation goes as follows:

We consider the given system $A$, immersed in a very large heat reservoir $A'$ [...] On attaining a state of mutual equilibrium, the system and the reservoir would have a common temperature, say $T$. Their energies, however, would be variable and, in principle, could have, at any time t, values lying anywhere between $0$ and $E^{(0)},$ where $E^{(0)}$ denotes the energy of the composite system [...] $$E_r + E'_r=E^{(0)}=\rm const.$$ ... Let the number of these states be denoted by $\Omega'(E'_r)$ [...] $$P_r \propto \Omega'(E'_r) \equiv \Omega'(E^{(0)} -E_r).$$ [...] we may carry out an expansion [...] around $E_r =0.$ However, for reasons of convergence, it is essential to effect the expansion of the logarithm instead: $$\ln \Omega'(E'_r) = \ln \Omega'(E^{(0)}) + \left(\frac{\partial \ln \Omega'}{\partial E'} \right)_{E'=E^{(0)}}(E'_r - E^{(0)}) + ...$$ $$\approx const - \beta'E_r,$$ [...] in equilibrium, $\beta' = \beta = 1/kT.$$ [...] $$P_r \propto \exp(-\beta E_r)$$

I have a few questions here:

1. For many systems, is it not the case that the temperature can be expressed as a function of the energy? An ideal gas is one such example. In that case, supposing I had an ideal gas inside a reservoir, which was free to exchange energy, but held at fixed particle number and volume, how could it be possible for it to be at a common temperature, but still take on any energy value?
2. Why is it essential to effect the expansion of the logarithm instead? What is the justification for this?
3. (Edit:) Why is the relevant probability taken to be $$P_r \propto \Omega' (E'_r)$$ rather than $$P_r \propto \Omega' (E'_r) \cdot \Omega(E_r)?$$ Thanks for any help.

Answer 1

For many systems, is it not the case that the temperature can be expressed as a function of the energy? An ideal gas is one such example.

What you're referring is the relationship between the internal energy and temperature. The internal energy is a thermodynamic quantity of the system which is calculated as an ensemble average from Statistical Mechanics. What Pathria here refers to here is not the ensemble average energy, it is simply the energy of the system at any instant in time. Ensemble averaged energy (or internal energy) would be equivalent to time-averaged energy, according to ergodic hypothesis. Even in the case of ideal gas, the internal energy is related to the temperature as $U \propto k_B T$, not the instantaneous energy.

In that case, supposing I had an ideal gas inside a reservoir, which was free to exchange energy, but held at fixed particle number and volume, how could it be possible for it to be at a common temperature, but still take on any energy value?

This is precisely the framework of canonical ensemble in statistical mechanics. The system is free to exchange energy and as a result a thermal equilibrium of common temperature is attained between the reservoir and system. To physically understand this, imagine the system on an average has energy $U$, but the system's energy changes instantaneously, fluctuating around the value $U$. While the energy fluctuates around $U$, the temeprature $T$ of the system and reservoir remain constant, such that the temperature is directly related to the average energy $U$.

Why is it essential to effect the expansion of the logarithm instead? What is the justification for this?

$\Omega(E)$ scales as the density of states of the system, which grows as $E^N$, where $N$ is the number of particles in the system. Realistically speaking, this is a extremely sharply rising function as $N$ ~ $10^{23}$. Therefore taking logarithmic expansion gives a much better approximation.