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Thermodynamics help 
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#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: 1,969

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

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


#4
Sep3013, 10:58 AM

P: 75

Thermodynamics help



#5
Sep3013, 11:00 AM

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#6
Sep3013, 11:08 AM

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PF Gold
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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

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#8
Sep3013, 11:23 AM

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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

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#10
Sep3013, 11:45 AM

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#11
Sep3013, 12:36 PM

P: 75

I'm not sure what to do with that 


#12
Sep3013, 01:31 PM

P: 1,969

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

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PF Gold
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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: 1,969

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

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PF Gold
P: 1,325

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


#16
Sep3013, 03:26 PM

P: 1,969

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

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[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: 1,969

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. 


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