Adiabatic process and heat exchange

[Vegeta]

All the places I've searched I find that the change in the internal energy for the adiabtic process is

dU = C_V\,dT

But I don't understand why they use C_V, neither the volume nor the preassure is constant during the adiabtic process.. I know that

dU = C_V\,dT = C_p\,dT - p\,dV

But why can that expression for dU be used for the internal energy. In the book "Physics for Scientists and Engineers", they say that because U is only a function of temperatur, then

U(T)=\int_0^{T} C_V\,dT

Without explaining why.. I hope someone can explain this. Thx in advance.

 

[Physics Monkey]

Adiabatic processes are those in which no heat is exchanged (entropy is constant), thus the change in internal energy is given by dU = - p \,dV from the first law. Now, as you correctly indicate, neither the pressure, temperature, or volume is constant in an adiabatic process, but the combination p V^\gamma is. This condition, together with the ideal gas law, enables you to integrate the above expression.

From an entirely different point of view, the internal energy is a state function and for an ideal gas it depends only on temperature. This means that irrespective of any other considerations, the change in internal energy for any process starting at T_i and ending at T_f is simply given by \Delta U = \int^{T_f}_{T_i} C_V \,dT.

Hope this helps.

 

[Vegeta]

I'm familiar with all the things you said. So I don't think that answers my question, which is why can the change in internal energy be expressed like

\Delta U = \int^{T_f}_{T_i} C_V \,dT

by using the heat kapacitet for constant volume C_V.
In another forum, a guy says, that because

C=\frac{\partial Q}{\partial T}=\frac{\partial U}{\partial T}+p\,\frac{\partial V}{\partial T}

Then for constant volume

C_V=\frac{\partial U}{\partial T}

And he says, that this generel, and doesn't depend on the state of the system. But with an adiabatic process you can't say that

p\,\frac{\partial V}{\partial T} = 0

Or am I completely wrong?

 

[Vegeta]

Oh, and by the way, I want to know, because I want to prove p V^\gamma, so I can't use this as you say.

 

[Physics Monkey]

From the first law, the heat capacity at constant volume is just \frac{\partial U}{\partial T}, that's all. The other guy was wrong, or at least misleading, "heat" isn't a state function so you can't take partial derivatives of it. Now, the fact that the heat capacity at constant volume doesn't relate to the heat exchanged (none) in an adiabatic process is irrelevant, it is still always the partial derivative of internal energy with respect to temperature. Therefore, when the internal energy depends on temperature alone, you can always use the heat capacity at constant volume to calculate the change in internal energy.

 

[Vegeta]

Ahh ok now I see. Then for constant preassure,

\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_p

And because C_p = C_V + nR and that p\,\frac{\partial V}{\partial T}=nR (is the last expression correct?), then

\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_V+nR = C_V

But again this is proved by using an isobaric and isochoric process, what is the argument that this will hold, if neither the volumen nor the preassure are kept constant (adiabatic)?

 

[Physics Monkey]

Yes, at constant pressure, the ideal gas law clearly states that p \frac{\partial V}{\partial T} = n R.

You can also get the result even more directly, consider the first law for a constant volume process: dU = \delta Q. By definition, \delta Q = C_V\, dT at constant volume so clearly C_V = \frac{\partial U}{\partial T}. This is the definition of C_V so it must always be true, and there really isn't much more to it than this. The internal energy is a state function and so is its partial derivative with respect to temperature. Now, you can regard your work above as a proof of the statement that C_P = C_V + nR for constant pressure processes, but since C_P and C_V are both state functions, this equation must actually hold for all processes.

 

[Vegeta]

So by showing that

C_V = \frac{\partial U}{\partial T}

Holds for constant preassure and constant volume, it is argument enough to say that it most hold, even if both of those thermodynamic variables p and V changes? That is for an adiabatic process.

 

[Physics Monkey]

Yes, C_V is a state function, and once you show that two state functions are equal, no matter what kind of process you used to show it, the equality is valid in general.

 

[Andrew Mason]

Adiabatic processes are those in which no heat is exchanged (entropy is constant),

Be careful here. Entropy change is not 0 for all adiabatic processes. Just reversible ones.

AM

 

[Physics Monkey]

AM,

We were clearly talking in the context of the quasi-static limit here, in this case the two are one and the same.

 

[tensus2000]

Hey dudes,
quick question, and I'm basically to lazy to figure out right now,
In an Otto cycle PV diagram what is held constant in the adiabatic processes, tempreture?
dunno,
odviously volume for isochors