# Entropy of Mixing for Two Ideal Gases at Different Temperatures

by Dahaka14
Tags: entropy, gases, ideal, mixing, temperatures
 P: 73 1. The problem statement, all variables and given/known data Two monatomic ideal gases are separated in a container by an impermeable wall, with volumes $V_{1}$ and $V_{2}$, temperatures $T_{1}$ and $T_{2}$, number of atoms $N_{1}$ and $N_{2}$, and both are at the same, constant pressure $P$. The wall is then removed, and the pressure is continued to be held constant. Calculate the change in entropy of this event. 2. Relevant equations The thermodynamic identity: $dU = \tau d\sigma - P dV + \mu dN$ where $U$ is the internal energy, $\mu$ is the chemical potential, $\tau$ is the temperature in fundamental units, and $\sigma$ is the entropy. Ideal gas equation: $P V = N \tau$ Average thermal energy for a monatomic ideal gas: $U = \frac{3}{2} N \tau$ 3. The attempt at a solution Examine the change in entropy of each gas, and then add the two changes together to get the total change. Since each gas will have the same number of particles after the change, the differential change in $U$ for each gas will be $dU = \tau d\sigma - P dV$. Rearranging to find the differential change in entropy, $d\sigma = \frac{dU + P dV}{\tau}$. Using the average thermal energy of a monatomic ideal gas, $U = \frac{3}{2} N \tau \implies dU = \frac{3}{2} N d\tau$, and the ideal gas equation, $P V = N \tau \implies V = \frac{N \tau}{P} \implies dV = \frac{N d\tau}{P}$, and substituting these relations, we get $d\sigma = \frac{\frac{3}{2} N d\tau + P \frac{N d\tau}{P}}{\tau} = \frac{\frac{5}{2} N d\tau}{\tau}$. Integrating both sides, we get $\int_{\sigma_{i}}^{\sigma_{f}} d\sigma = \int_{\tau_{i}}^{\tau_{f}} \frac{\frac{5}{2} N d\tau}{\tau} \implies \Delta \sigma = \frac{5}{2} N \log \left( \frac{\tau_{f}}{\tau_{i}} \right)$. Now we add the corresponding expressions for each gas to get the total entropy change: $\Delta \sigma_{1} + \Delta \sigma_{2} = \frac{5}{2} N_{1} \log \left( \frac{\tau_{f}}{\tau_{1}} \right) + \frac{5}{2} N_{2} \log \left( \frac{\tau_{f}}{\tau_{2}} \right)$ I am a little uneasy about this solution and how to further simplify this since thermodynamics and statistical mechanics are my weakest areas.First of all, is this correct so far? One question I have at this point is: why is there no expression for volume involved at the end? I know that for the entropy of mixing of an analogous problem, only with identical initial temperatures, involves a solution containing the initial and final volumes of each gas. Thus, should we not be adding an additional term to that solution to produce a larger change in entropy? Second, how do I know what the actual final temperature is if we are not aware of the different gases involved?
 HW Helper Thanks PF Gold P: 4,408 I have to say I'm confused about the "fundamental" unit of temperature, which I guess for you is kT, T in Kelvin. OK, maybe that's cool, but then is your first equation dimensionally consistent? Looks like U is what I understand to be U, but then shouldn't your tau be T? Or is your σ not really entropy but entropy in "fundamental units"? Does tau*dσ = TdS I hope? I'm willing to look at this some more once my question is cleared up.
 HW Helper Thanks PF Gold P: 4,408 OK, never mind. I will just use what I'm used to. Let pij = pressure of gas i in state j, i, j = 1 or 2 Let Tf = final temperature Let ni = no. of moles of gas i R = 8.317 J/mole-K p11V1 = n1RT1 p22V2 = n2RT2 (p12 + p22) = (n1 + n2)RTf or Tf = p/(n1 + n2)R For ideal gas, dS = Cp*dT/T - nR*dp/p where Cp = heat capacity at constant p But, since dp = 0, dS1 = Cp1*dT/T dS2 = Cp2*dT/T dS = dS1 + dS2 ΔS = ∫dS1 from T1 to Tf + ∫dS2 from T2 to Tf ΔS = Cp1*ln(Tf/T1) + Cp2*ln(Tf/T2) This looks very much like what you got. I leave it to you to change the n's to N's. I notice you assume Cp = 5R/2 but this as you know is an approximation for a monatomic gas. There is a constant term to be added, separately for each gas, to get a more or less accurate value for Cp.

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