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Very, very small. Today, the temperature of the Universe is $T_0 = 2.73\,\mathrm{K}$, so when the temperature was $T = 300\,\mathrm{K}$, the redshift was $$ z = \frac{T}{T_0} - 1 \simeq 100, $$ which means that the Universe was only $t = 15\,\mathrm{Myr}$ old, long before the first stars were born. However, this was still long after the Universe went from ...


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The CMB has a temperature of $T=2.73(1+z)$. So when it was at 300K, the redshift $z=110$. At this redshift there were no stars (or at least for the purposes of this question, there is no significant reionisation, which happened at $z\leq 20$, so we can assume the universe is neutral H and He atoms. The universe at that time would have been matter dominated (...


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It was during the period called the dark ages. Between 3000K and 60K the Stars had not formed, thus dark. Times between 380000 years to about 150 million years after the Big Bang. See the chronology at https://en.m.wikipedia.org/wiki/Chronology_of_the_universe 3000K was roughly recombination, and then atoms started forming. For most of that time it was ...


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"The sum of the entropy variations is 00 since the system is isolated". If by entropy variation you mean total change in entropy (as you do indeed), then it is not true. Equalization of temperature and pressure have taken place, which are irreversible processes. So entropy has increased and reached a maximum value consistent with constraints on your system (...


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The mean free path of gas molecules is $$\lambda =\frac{RT}{\sqrt 2 \pi d^2 N_A P}$$ Where $R$ is the gas constant, $d$ the diameter of the molecule and $N_A$ Avogadro's constant. The average relative velocity of gas molecules can be obtained by the Maxwell-Boltzmann distribution and is equal to $$\langle v \rangle = \sqrt{\frac{8kT}{\pi m}}$$ The mean ...


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Equilibrium thermodynamics should in principle only deal with equilibrium states. In an equilibrium state, the thermodynamic variables (for example $P,V,T$) are by definition constants: this is why the time variable $t$ never appears in thermodynamics. However, when we use an equation of state $f(P,V,T)=0$ such as the ideal gas equation of state, $PV=nRT$, ...


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This question is quite profound, but I didn't think it would go that far. I'll post my views for feedback, and since I'll write some lines it is best to leave it as an answer. As I mentioned briefly in my first comment, time derivatives won't make sense in thermodynamics. I'll try to develop this comment a little bit more. The entities we are talking ...


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If a fluid system is experiencing a transient irreversible change, the temperature and pressure within the system will typically be non-uniform spatially. This is the result of two factors not present in (slow) reversible changes: the inertia (mass) of the fluid (which allows non-uniformities in velocity and density) and the viscous nature of the fluid (...


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For quasi-stationary processes you can take the time derivative, no problem. Apart from the "state equation", there is a "process equation", for example, for an adiabatic expansion $PV^{\gamma}=\rm{const}$ or so. It is also quasi-stationary. What you can loose in taking the time derivatives are quick processes like sound or shock wave propagation and their ...



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