
#1
Sep2913, 09:07 PM

P: 75

1. The problem statement, all variables and given/known data
The question is: Consider a hypothetical ideal gas with internal energy U = NkT_{o}(T/T_{0})^{α+1}, where T_{o} and α are positive constants. Show that in an adiabatic process, V*exp[(1+1/α)(T/T_{o})^{α}] = constant. 2. Relevant equations PV^{γ} = constant γ = C_{p}/C_{v} C_{p} = C_{v} + Nk 3. The attempt at a solution I'm pretty sure that I'm supposed to show that [(1+1/α)(T/T_{o})^{α}] is equal to γ and since PV^{γ} = constant, V*exp[(1+1/α)(T/T_{o})^{α}] = constant. When I try to solve it though I can't get the solution to come out. I differentiate U to get C_{v} = Nk(1+α)(T/T_{o})^{1+α}. When I plug that into γ I get γ = 1 + 1/[(1+α)(T/To)^{α}]. Either I'm just not simplifying it enough and the answer is correct, or I solved for λ incorrectly, or my equations are incorrect. I don't know which it is though. 



#2
Sep2913, 10:06 PM

P: 1,902

PV^{γ} = constant is not valid for this problem. It follows from the usual internal energy for ideal gas, U(T)=nCvT Here you have a different function U(T) and you have to find the relationship between volume and temperature. (for the "usual" ideal gas this will be TV^{γ1} = constant ) You can use the first law in conjunction with the equation of state to do this. 



#3
Sep2913, 10:32 PM

HW Helper
Thanks
P: 9,817

Is PV^{γ} = constant when Cv depends on T? ehild 



#4
Sep3013, 10:58 AM

P: 75

Thermodynamics help 



#5
Sep3013, 11:00 AM

P: 75





#6
Sep3013, 11:08 AM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,420

dU=C_{v}dT PV=RT From the first law, for an adiabatic reversible process, how is dU related to PdV? 



#7
Sep3013, 11:18 AM

P: 75





#8
Sep3013, 11:23 AM

HW Helper
Thanks
P: 9,817

U is given as function of T. PV=kNT is valid for the ideal gas, and also the First Law is valid. For an adiabatic process dU=PdV. Use P=kNT/V, and integrate.
ehild 



#9
Sep3013, 11:31 AM

P: 75





#10
Sep3013, 11:45 AM

Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,154





#11
Sep3013, 12:36 PM

P: 75

I'm not sure what to do with that 



#12
Sep3013, 01:31 PM

P: 1,902

But the idea is "forget gamma". And "forget pv^gamma". Does not apply here. 1. From first law applied to adiabatic process you have: dU=pdV You have U(T) so find dU. 2. You have PV=nRT so you can eliminate p on the right hand side: pdV= nRTdV/V So you will have an equation relating V and T. Integrate (after separating variables) and you'll find that exponential relationship. 



#13
Sep3013, 01:57 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,420

You have the equation of U as a function of T, and you know know that
[tex]C_v=\frac{\partial U}{\partial T}[/tex] Just differentiate the equation for U with respect to T, and write [tex]dU=C_vdT=\frac{\partial U}{\partial T}dT=PdV[/tex] Then, just substitute the ideal gas law for P, and integrate. 



#14
Sep3013, 02:14 PM

P: 1,902

Cv=∂U/∂T is valid for a "real" ideal gas. The whole point here is that U(T) is not given by dU=CvdT but by that other, more complicated formula. If he does what you suggest he'l get just the usual [tex]TV^{\gamma 1 }= constant[/tex] and not the formula required by the problem. But the method will work. This is what I tried to explain as well. Just use [tex]dU=Nk(\alpha +1) (T/T_0)^{\alpha} dT[/tex]. There is no need to introduce Cv or gamma. 



#15
Sep3013, 02:54 PM

Sci Advisor
PF Gold
P: 1,111

$$ C = \frac{Q}{\Delta T} $$ by considering a constant volume (hence ##W=0##), without invoking an ideal gas. 



#16
Sep3013, 03:26 PM

P: 1,902

Oh, I see. I used partial derivatives. I meant that dU=CvdT may not apply to other systems. It is valid only for some systems, like ideal gas in the "proper" definition. So dU=Nk(α+1)(T/T0)^α dT You don't need to define or use a specific heat to solve the problem. Sorry for the confusion. 



#17
Sep3013, 07:02 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,420

[tex]C_v=Nk(\alpha +1) (T/T_0)^{\alpha}[/tex] Are you uncomfortable with an ideal gas heat capacity that varies with temperature. A temperaturedependent heat capacity is part of the definition of an ideal gas that we engineers use. 



#18
Sep3013, 08:38 PM

P: 1,902

I don't feel any discomfort about temperature variation of Cv or about Cv in general. Even Cp it's bearable, despite all these pressure variations. 


Register to reply 
Related Discussions  
Thermodynamics Zemansky Heat and Thermodynamics book question  Advanced Physics Homework  0  
Thermodynamics and Statistical Thermodynamics References  Science & Math Textbook Listings  9  
Physical Chemistry books (Thermodynamics/Statistical Thermodynamics/Kinetics/QM)  Science & Math Textbook Listings  1  
Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engine  Introductory Physics Homework  7  
Prof teaches Statistical thermodynamics in a Classical Thermodynamics class  Academic Guidance  10 