Measurement of observables of a gas in a weakly interacting room Suppose I have a gas at equilibrium in a box weakly interacting with the room (weak exchange of energy but absolutely no exchange of particles, so that we are in a canonical ensamble). The gas interacts with the room. Does this count as a measurement? In what eigenstate does the gas collapse to then? The room has the same temperature of the gas, which makes me think the gas should collapse to some weird superposition of energy eigenstates with one of them with a much higher coefficient in the energy eigenstates expansion. Are the coefficient of the superposition related to the Boltzmann distribution $ \frac{1}{Z} e^{-\beta E_m} $? If so what are the phases of the expansion? From what I understood from my book the gas should be in a single energy eigenstate (with probability given by the Boltzmann distribution) but that doesn't make any sense! If the gas really were in a single energy eigenstate it wouldn't evolve in time. While taking a measurement of an observable it wouldn't make sense to calculate the average value of the observable over every energy eigenstate and then weight it with Boltzmann weights $ \frac{1}{Z} e^{-\beta E_m} $ in the average because the "real state" would only be one of them and not a superposition of them all. 
 A:  Summary 
If we treat the room as a boundary condition (like a typical particle-in-a-box homework problem), then no: the energy of the gas is not measured in such a model. 
If we treat the room as a quantum system made of molecules (like a real physical room), then the gas becomes entangled with the room. The entanglement is practically irreversible and remains approximately diagonal in the energy basis, so this does count as a measurement of the energy of the gas. However, as explained below, the measurement has finite resolution (in energy), so the gas is not in a stationary state. I'll show that the resolution of the energy measurement is coarse enough that the gas can still evolve rapidly in time.
 The gas still evolves in time 
First, here's a qualitative explanation of why a gas in equilibrium is not in a stationary state.
Statistical mechanics works by counting states. In classical statistical mechanics, we count the number of microstates compatible with specified macroscopic conditions. In quantum statistical mechanics, we count the number of dimensions of the part of the Hilbert space that is compatible with those conditions. The state-counting method leads to the conclusion that the gas can be described by a density matrix of the form $\rho\propto\exp(-\beta H)$, where $H$ is the Hamiltonian of the gas. This state is time-independent, which leads to the paradox in the question.
The paradox is resolved by remembering that the state-counting method ignores the interaction between the gas and the room. In the state-counting method, the dynamics is replaced by the assumption that all microstates of the combined system are equally likely, which is easier to handle mathematically. In reality, the gas and the room do interact with each other. If we start with some specific initial pure microstate $|\psi\rangle$ of the combined system (gas + room), we would find that the reduced density matrix of the gas actually does depend on time, fluctuating slightly about the nominal time-independent state $\propto \exp(-\beta H)$ that was deduced using the easier state-counting method. This occurs as a result of the ongoing interactions between the gas and the room, even if we contrived the initial reduced density matrix to be exactly $\propto \exp(-\beta H)$.
Calculations that explicitly account for the interaction are difficult, but for the purpose of the question, we can quantify the effect simply in terms of the resolution $\Delta E$ of the energy measurement, which is estimated below. To see why this is sufficient, remember that the state-counting method counts the number of dimensions of the part of the Hilbert space that is compatible with the given macroscopic conditions. The gas is equally likely to be in any one of those states, most of which are not stationary states. In particular, if that part of the Hilbert space spans a range $\Delta E$ of energies, then the gas can be in any superposition of energy eigenstates having energies in this range. Such a state can evolve into an orthogonal state on a timescale of order $\Delta t\sim \hbar/\Delta E$. The goal is to estimate the timescale $\Delta t$, which we can do by estimating the resolution $\Delta E$ of the energy measurement.
 The resolution of the energy measurement 
On paper, if we measure the energy $E$ of the gas perfectly a jillion times and make a histogram of the results, the histogram would be proportional to
$$
 n(E)\propto e^{S(E)} e^{-\beta E}
\tag{1}
$$
where $S(E)$ is the entropy function. This is just the density of states $e^{S(E)}$ times the Boltzmann distribution.
This histogram is dominated by a very sharp peak at the specific value of $E$ that satisfies $dS/dE=\beta$, but even though the peak is very sharp, it still has a finite width $\Delta E$. This width characterizes the resolution of the natural energy measurement that occurs due to the interaction between the gas and the room. 
To estimate $\Delta E$, consider an ideal gas with $N$ molecules in a fixed volume. In this case, the entropy function is $S(E)=\frac{3N}{2}\log E$, which gives
$$
 n(E)\propto \exp\left(\frac{3N}{2}\log E - \beta E\right).
\tag{2}
$$
Write the energy as $E=E_0+\epsilon$ where $E_0$ is the energy that maximizes $n(E)$, and expand (2) to second order in $\epsilon$. The peak is so sharp that this should be an excellent approximation. The result is
$$
 n(E_0+\epsilon)\propto n(E_0)\exp\left(
 -\frac{\beta^2\epsilon^2}{3N}+O(\epsilon^3)\right),
\tag{3}
$$
which shows that the width of the peak is 
$$
 \Delta E\sim \frac{\sqrt{N}}{\beta}.
\tag{4}
$$
Let's plug in some realistic numbers: for the number of molecules in the gas, use $N\sim 10^{24}$ (Avagadro's number), and for the temperature, use room temperature $T\sim 300$ K. Boltzmann's constant is $k\sim 10^{-23}$ J/K, and $\beta=1/kT$, so the width is
$$
 \Delta E\sim kT\sqrt{N}\sim 10^{-9}\text{ J}. 
\tag{5}
$$
The corresponding timescale is
$$
 \Delta t\sim\frac{\hbar}{\Delta E}\sim 10^{-25}\text{ seconds}.
\tag{6}
$$
Altogether, this says that if the gas has roughly Avagadro's number of molecules, then the room measures the energy of the gas with a resolution of $\sim 10^{-9}$ Joules, so the gas can still evolve in time on a scale of $10^{-25}$ seconds, which is fast enough to resolve the paradox. A gas with more molecules will evolve even faster.
