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An insulated container (constant volume, adiabatic) contains an ideal gas.

We open the container's hatch for a few seconds and let some particles escape from the container, then we close the hatch. We know container's pressure has reduced by exiting particles. What can we say about gas temperature?

I have got the approach below, I would like to get your comments on this, please.

enter image description here

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  • $\begingroup$ I think $T2<T1$ , as with gay lussac's law at constant volume Temperature varies directly with pressure for an ideal gas , If $P_{2}<P_{1}$ is correct Then $T_{2}<T_{1}$ Must also do the same. Intuitively the lower Kinetic energy of particles therefore lower the absolute value of temperature makes sense to me $\endgroup$
    – Naveen V
    Mar 9 at 16:59
  • $\begingroup$ @Navaeen V, according to ideal gas state formula, we have less mole of gas (smaller n) and lower pressure, just by looking at the odeal gas formula (PV=nRT), one cannot say temperature must have changed. $\endgroup$
    – Ebi
    Mar 10 at 5:25

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This is an unsteady state process and the answer depends on the precise details of the process. If instead of opening the hatch we vent the gas via a valve, we may treat this process as isenthalpic (throttling). With the further assumption that the contents of the tank undergo isentropic change, which amounts to neglecting irreversibilities due to flow inside the tank, we obtain two equations that fix the state of the system inside the tank: $$ S(P,T) = S(P_0,T_0),\quad H(T,P) = H(T',P_\text{ext}) $$ where $(P,T)$ are the conditions inside the tank after the venting, $(P_0,T_0)$ are the conditions before venting, $P_\text{ext}$ is the pressure to which the gas vents and $T'$ is the temperature of the gas immediately after venting. In writing these equations we are assuming that the gas that is vented is at the pressure of the surroundings but at temperature $T'$ determined by the isenthalpic condition. For an ideal gas, $T'=T$.

The bottom line is that one cannot solve this problem without making enough assumptions to deal with the complicated nature of the process.

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  • $\begingroup$ thank you. I edited my question and added some calculations showing T2<T1. What's your verdict on that? $\endgroup$
    – Ebi
    Mar 12 at 1:35
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    $\begingroup$ Yes, this is correct. The final temperature is indeed lower but the reason is not that the hotter molecules escape leaving behind the colder ones. It is the expansion that causes the cooling. Maybe you can edit that part of the question. $\endgroup$
    – Themis
    Mar 12 at 19:33

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