I see answers to similar questions like this and this that confuse me. I feel like they ignore how the expansion of the gases occurs.

Consider two chambers of gases at $ n_{1i}, P_{1i} $ and $ n_{2i}, P_{2i} $ each with constant initial temperature $ T_i $. Put them in adiabatic contact such that the total volume $ V = V_1 + V_2 $ is constant, no heat exchange $ dT_i = 0 $, and the chambers are allowed to expand against each other based on their pressures. Final pressure will be equilibrium $ P_f $

As they expand, they should be subject to Adiabatic relation for P-T,

$$ P_{1i}^{1-\gamma} T_i^\gamma = P_f^{1-\gamma} T_{1f}^\gamma \\ P_{2i}^{1-\gamma} T_i^\gamma = P_f^{1-\gamma} T_{2f}^\gamma \\ $$

Dividing those, you can get a relation, $$ T_{1f} = \left(\frac{P_{1i}}{P_{2i}}\right)^{(1 - \gamma)/\gamma} T_{2f} $$

Only now do gas mixing according to the formulas in the linked posts,

$$\begin{align} T_{mixed} &= \frac{n_1 \left(\frac{P_{1i}}{P_{2i}}\right)^{(1 - \gamma)/\gamma} + n_2}{n_1 + n_2} T_{2f} \\ &= \frac{n_1 \left(\frac{P_{1i}}{P_f}\right)^{(1 - \gamma)/\gamma} + n_2 \left(\frac{P_{2i}}{P_f}\right)^{(1 - \gamma)/\gamma}}{n_1 + n_2} T_{i} \\ \end{align}$$

$ \gamma = 7/5 $ for diatomic, so $ (1 - \gamma)/\gamma < 0 $ implying that we would expect cooling $ T_{mixed} < T_i $ if combining a low pressure gas with a high pressure gas ($ P_{2i} > P_{1i} $) in equal volumes (implying $ n_2 > n_1 $) which intuitively makes sense.

Am I wrong? If I am correct, then this would imply that this is either incorrect or incomplete because they predict no temperature change.

  • 1
    $\begingroup$ The adiabatic relation for P-T you are using is for an ideal gas undergoing a reversible adiabatic (constant entropy) process. Allowing the gases to spontaneously expand against one another is not a reversible process $\endgroup$
    – Bob D
    Dec 11, 2020 at 10:49
  • $\begingroup$ @BobD Can you explain why it isn't reversible? Can this fact be proven? $\endgroup$
    – ignorance
    Dec 11, 2020 at 13:54
  • $\begingroup$ To be clear, I am considering a case like physicsforums.com/threads/…. I feel like there's a lot of qualitative descriptions, but it's not clear to me what the actual dynamics are. $\endgroup$
    – ignorance
    Dec 11, 2020 at 14:20
  • 1
    $\begingroup$ Any unrestrained expansion process is irreversible. That is because during the expansion the gas in not in internal equilibrium (temperature and pressure gradients abound). To be reversible the expansion has to be carried out very slowly so that the g. as is always in internal equilibrium. So unless in your example there is some mechanism for controlling the expansion process, the process must be considered irreversible. $\endgroup$
    – Bob D
    Dec 11, 2020 at 14:28


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.