Thermodynamics - adiabatic process 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.
The equation 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^{γ-1} = constant ) You can use the first law in conjunction with the equation of state to do this.
For an ideal gas, dU=C_{v}dT PV=RT From the first law, for an adiabatic reversible process, how is dU related to PdV?
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
No, that would not follow, so I don't think you want to show that [(1+1/α)(T/T_{o})^{α}] is equal to γ. Instead, try raising V*exp[(1+1/α)(T/T_{o})^{α}] to the power of γ, using the expression for γ that you derived.
Well, I don't understand what part you don't understand. 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.
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.
This is a "hypothetical" ideal gas. 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.
I have to disagree. Cv=∂U/∂T follows simply from the definition of heat capacity, $$ C = \frac{Q}{\Delta T} $$ by considering a constant volume (hence ##W=0##), without invoking an ideal gas.
Did I say anything that seem to contradict your statement? I just meant just that you don't need Cv to solve the problem. It does not appear in this problem. 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.
This is exactly what I was suggesting. I brought the heat capacity into the picture because I felt the OP would feel more comfortable with it. For this particular ideal gas, C_{v} is not independent of temperature, but is given by: [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 temperature-dependent heat capacity is part of the definition of an ideal gas that we engineers use.
Is this a question? 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.
Oops. I left out the question mark. Thank you for serving as the grammar police enforcer. Getting back to the thread, I think we are (and were) totally in agreement on how this problem should be solved. Of course, for an ideal gas, C_{p} is also a function only of temperature. Chet