Ideal gas and dimensionless entropy.

[ted1986]

Hi,

I'm trying to solve the attached exercise, but I'm not sure I'm in the right direction.
What I've been trying to do is using the relation:
dU=TdS-pdV
Dividing it by T and adding dT so the relation will be
(dU/T)dT=(ds/dT)-pT(dVdT), but I don’t think it’s the right way to solve it.

Can someone help me?

thnks.

 

[Ken G]

Start with your expression for dU, and solve it for dS. Then figure out how changes in rho, called d rho, depend on changes in V, called dV. Now figure out how U, the energy, depends on T, V, and rho, which requires you to look up formulas for how both gas and radiation energy depend on these quantities. Once you have that, add the two U contributions together, factor out the gas contribution (that's where you will get the sigma coming in), and find dU.

 

[Ken G]

By the way, it will help you quite a bit to know that:
U = (3 P_r + 3/2 P_g) V
so
dU = (3 dP_r + 3/2 dP_g) V + (3 P_r + 3/2 P_g) dV
and then you will also need the ideal gas law for P_g and the fact that
dP_r/P_r = 4 dT / T.
Do you know things like:
drho / rho = - dV / V ? That's the kind of expression you'll need to understand well.
You have yourself a challenging problem, try to find expressions like dT/T and drho/rho, and factor out a P_g when you get to the end, that's where sigma comes in. I think you will find this exercise pretty challenging, but it contains a lot of the skills you need now.

 

[ted1986]

By the way, it will help you quite a bit to know that:
U = (3 P_r + 3/2 P_g) V
so
dU = (3 dP_r + 3/2 dP_g) V + (3 P_r + 3/2 P_g) dV
and then you will also need the ideal gas law for P_g and the fact that
dP_r/P_r = 4 dT / T.
Do you know things like:
drho / rho = - dV / V ? That's the kind of expression you'll need to understand well.
You have yourself a challenging problem, try to find expressions like dT/T and drho/rho, and factor out a P_g when you get to the end, that's where sigma comes in. I think you will find this exercise pretty challenging, but it contains a lot of the skills you need now.

Thank you for your answer.
I used the relations you wrote, and it looks like I'm in the right direction, but as you can see in the attached file – I don’t know how to use the entropy dS instead of dU.

B.T.W - Where can I find explanations about the relations you wrote?
Thank you.

 

[Ken G]

Yes, that's looking pretty good-- now solve your final expression for dS and look for a way to replace some stuff with the Boltzmann k (which you then absorb into dS because they are using the dimensionless entropy and we are not). The place to look for these kinds of relations is in regular gas dynamics textbooks, it is what underlies the ideal gas laws.

 

[ted1986]

Yes, that's looking pretty good-- now solve your final expression for dS and look for a way to replace some stuff with the Boltzmann k (which you then absorb into dS because they are using the dimensionless entropy and we are not). The place to look for these kinds of relations is in regular gas dynamics textbooks, it is what underlies the ideal gas laws.

OK, I expressed dS, but it looks different from the final term (File is attached).
Another question - the p (from the relation TdS-pdV=dU) - is it the total pressure? and if it is, how is it expressed?

 

[Ken G]

You're very close. Note that dividing through by nR (or Nk, which is the same thing) is what turns the total dS into a dimensionless dS per particle (which is what they mean by dS). In P dV, the P is the total pressure, so P_g + P_r. That should make sense to you-- the work done doesn't care what flavor of pressure is doing it, and the thermodynamic equilibrium insures that radiation and gas will share the energy to be at the same T.

 

[ted1986]

Solved!

Thank you so much!