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The gas is expanded adiabatically and then isothermally. Thus the temperature it has at the end of adiabatic expansion stays the same even after the isothermal process. Ideal Gas equation after adiabatic expansion: $p_aV_a=nRT_a$, where index "a" shows after. You do not have $V_a, T_a$ in this equation. However, another equation you can write down is the ...

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This is mentioned, for example, in "Statistical Mechanics: Algorithms and Computations" by Krauth section 2.1.2. (But the author just uses computational/empirical details and you've gone into it more). So the result you get is correct and echoed elsewhere. The errors will blow up rapidly.

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Prahar is correct that generally we have an energy contribution of ${1 \over 2} kT$ per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of ${3 \over 2} kT$. Often people talk about thermal energy being '$kT$' because of the exponential expression in $N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}}$ ...

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The thermal energy of a system is $$E = f \frac{1}{2} k T$$ where $f$ is the number of degrees of freedom of the theory - which is roughly speaking the number of dimensions it is allowed to move in. For instance, if you are talking about an atom in 3 space dimensions, then the atom can move along the 3 axes and hence $f=3\implies E = \frac{3}{2} kT$. If ...

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There are two issues in this question, one is how to compute the "reduced" phase space density (i.e. the marginal distribution on the reduced phase space), and the other is how to deal with continuum phase spaces and distributions on it. Let's face the first problem, in a simplified situation. Suppose you have two sets $A = \{a_{1},a_{2}\}$ and ...

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Another way of looking at this is from the thermodynamic equilibrium point of view. Saying that a gas is at a certain temperature implies something called "local thermodynamic equilibrium" (LTE). The populations of atoms in their various energy levels or even ionisation states is determined by this temperature through the Boltzmann factors $\exp(-E/kT)$. The ...

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The first point is that a photon carries momentum, so anything that absorbs it acquires that momentum. So after absorbing a photon a gas molecule/atom will have an increased momentum. If the atom simply re-emits the photon then the momentum of the atom goes back to what it was before and nothing has changed. Heating occurs when the excited atom collides ...

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In one way, we can assume that particles repulsing when they "approach close by" is similar to particles with larger sizes colliding. It means that particles cannot approach arbitrarily close. This may not be exactly what you are distinguishing with "physical collision", but it has consequences for ideal gases. The greater the distance of this ...

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An ideal gas should consist of pointlike particles that are non-interacting, except if they collide, in which case they should do so elastically, without losing kinetic energy. I do not think there is any distinction here between a collision and a repulsive force. Any short-range repulsive force between particles (short-range compared with the average ...

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