Entropy of Mixing for Two Ideal Gases at Different Temperatures

In summary, we are calculating the change in entropy when two monatomic ideal gases are separated in a container by an impermeable wall and then the wall is removed, and the pressure is continued to be held constant. Using the thermodynamic identity and ideal gas equation, we can find the change in entropy for each gas and then add them together to get the total change. After integrating both sides, we get a final expression for the change in entropy that involves the initial and final temperatures for each gas. This solution is consistent with the use of fundamental units of temperature and the assumption of an ideal gas. However, for a more accurate calculation, the constant term for the heat capacity at constant pressure should be included for each gas.
  • #1
Dahaka14
73
0

Homework Statement


Two monatomic ideal gases are separated in a container by an impermeable wall, with volumes [itex]V_{1}[/itex] and [itex]V_{2}[/itex], temperatures [itex]T_{1}[/itex] and [itex]T_{2}[/itex], number of atoms [itex]N_{1}[/itex] and [itex]N_{2}[/itex], and both are at the same, constant pressure [itex]P[/itex]. The wall is then removed, and the pressure is continued to be held constant. Calculate the change in entropy of this event.

Homework Equations


The thermodynamic identity:
[itex]dU = \tau d\sigma - P dV + \mu dN[/itex]
where [itex]U[/itex] is the internal energy, [itex]\mu[/itex] is the chemical potential, [itex]\tau[/itex] is the temperature in fundamental units, and [itex]\sigma[/itex] is the entropy.

Ideal gas equation:
[itex]P V = N \tau[/itex]

Average thermal energy for a monatomic ideal gas:
[itex]U = \frac{3}{2} N \tau[/itex]

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 [itex]U[/itex] for each gas will be
[itex]dU = \tau d\sigma - P dV[/itex].
Rearranging to find the differential change in entropy,
[itex]d\sigma = \frac{dU + P dV}{\tau}[/itex].
Using the average thermal energy of a monatomic ideal gas,
[itex]U = \frac{3}{2} N \tau \implies dU = \frac{3}{2} N d\tau[/itex],
and the ideal gas equation,
[itex]P V = N \tau \implies V = \frac{N \tau}{P} \implies dV = \frac{N d\tau}{P}[/itex],
and substituting these relations, we get
[itex]d\sigma = \frac{\frac{3}{2} N d\tau + P \frac{N d\tau}{P}}{\tau} = \frac{\frac{5}{2} N d\tau}{\tau}[/itex].
Integrating both sides, we get
[itex]\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)[/itex].
Now we add the corresponding expressions for each gas to get the total entropy change:
[itex]\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)[/itex]

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?
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
Last edited:

What is entropy of mixing for two ideal gases at different temperatures?

Entropy of mixing for two ideal gases at different temperatures is a thermodynamic property that measures the randomness or disorder of the molecules when they are mixed together.

How is entropy of mixing calculated for two ideal gases at different temperatures?

The entropy of mixing can be calculated using the formula S = -nR(xAlnxA + xBlnxB), where n is the number of moles, R is the gas constant, and xA and xB are the mole fractions of the two gases.

Why does entropy of mixing increase when two ideal gases at different temperatures are mixed?

When two ideal gases at different temperatures are mixed, the molecules of each gas become more randomly distributed, leading to an increase in disorder or entropy. This is because the molecules now have more available microstates to occupy, increasing the overall randomness of the system.

How does the entropy of mixing for two ideal gases change with temperature?

The entropy of mixing for two ideal gases is directly proportional to the temperature difference between the two gases. As the temperature difference increases, so does the entropy of mixing.

Are there any factors that can affect the entropy of mixing for two ideal gases at different temperatures?

Yes, there are several factors that can affect the entropy of mixing for two ideal gases at different temperatures, such as the size and shape of the molecules, the pressure and volume of the mixture, and the presence of any chemical reactions or phase changes.

Similar threads

  • Advanced Physics Homework Help
Replies
4
Views
852
  • Advanced Physics Homework Help
Replies
1
Views
702
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
817
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
986
  • Advanced Physics Homework Help
Replies
5
Views
956
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
32
Views
971
  • Thermodynamics
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top