Adiabatic process and heat exchange

In summary, the change in internal energy for an adiabatic process can be expressed as dU = C_V\,dT, but the use of C_V may be confusing as neither volume nor pressure is constant. However, the ideal gas law and the condition pV^\gamma = constant allow for integration of the expression. Additionally, the internal energy is a state function and for an ideal gas it only depends on temperature, so the change in internal energy for any process can be calculated using the heat capacity at constant volume. This is true even for adiabatic processes where neither volume nor pressure is constant.
  • #1
Vegeta
22
0
All the places I've searched I find that the change in the internal energy for the adiabtic process is

[tex] dU = C_V\,dT [/tex]

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

[tex] dU = C_V\,dT = C_p\,dT - p\,dV [/tex]

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

[tex] U(T)=\int_0^{T} C_V\,dT [/tex]

Without explaining why.. I hope someone can explain this. Thx in advance.
 
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  • #2
Adiabatic processes are those in which no heat is exchanged (entropy is constant), thus the change in internal energy is given by [tex] dU = - p \,dV [/tex] from the first law. Now, as you correctly indicate, neither the pressure, temperature, or volume is constant in an adiabatic process, but the combination [tex] p V^\gamma [/tex] 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 [tex] T_i [/tex] and ending at [tex] T_f [/tex] is simply given by [tex] \Delta U = \int^{T_f}_{T_i} C_V \,dT [/tex].

Hope this helps.
 
  • #3
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

[tex] \Delta U = \int^{T_f}_{T_i} C_V \,dT [/tex]

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

[tex]C=\frac{\partial Q}{\partial T}=\frac{\partial U}{\partial T}+p\,\frac{\partial V}{\partial T}[/tex]

Then for constant volume

[tex]C_V=\frac{\partial U}{\partial T}[/tex]

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

[tex] p\,\frac{\partial V}{\partial T} = 0 [/tex]

Or am I completely wrong?
 
  • #4
Oh, and by the way, I want to know, because I want to prove [itex] p V^\gamma [/itex], so I can't use this as you say.
 
  • #5
From the first law, the heat capacity at constant volume is just [tex] \frac{\partial U}{\partial T} [/tex], 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.
 
  • #6
Ahh ok now I see. Then for constant preassure,

[tex]\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_p[/tex]

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

[tex]\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_V+nR = C_V[/tex]

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)?
 
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  • #7
Yes, at constant pressure, the ideal gas law clearly states that [tex] p \frac{\partial V}{\partial T} = n R [/tex].

You can also get the result even more directly, consider the first law for a constant volume process: [tex] dU = \delta Q [/tex]. By definition, [tex] \delta Q = C_V\, dT [/tex] at constant volume so clearly [tex] C_V = \frac{\partial U}{\partial T} [/tex]. This is the definition of [tex] C_V [/tex] 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 [tex] C_P = C_V + nR [/tex] for constant pressure processes, but since [tex] C_P [/tex] and [tex] C_V [/tex] are both state functions, this equation must actually hold for all processes.
 
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  • #8
So by showing that

[tex] C_V = \frac{\partial U}{\partial T} [/tex]

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.
 
  • #9
Yes, [tex] C_V [/tex] 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.
 
  • #10
Physics Monkey said:
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
 
  • #11
AM,

We were clearly talking in the context of the quasi-static limit here, in this case the two are one and the same.
 
  • #12
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
 

1. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no heat exchange between the system and its surroundings. This means that the system is insulated and there is no transfer of heat energy in or out of the system.

2. How does an adiabatic process differ from an isothermal process?

An adiabatic process differs from an isothermal process in that there is no heat transfer in an adiabatic process, while in an isothermal process, the temperature remains constant as heat is exchanged between the system and its surroundings.

3. What is the first law of thermodynamics and how does it relate to adiabatic processes?

The first law of thermodynamics states that energy can neither be created nor destroyed, only transferred or converted from one form to another. In an adiabatic process, there is no heat transfer, so any change in energy must come from work done on or by the system.

4. What are some real-world examples of adiabatic processes?

A common example of an adiabatic process is the compression or expansion of a gas in a closed container with insulated walls. Other examples include the expansion of a gas due to a sudden decrease in pressure, such as in a bicycle pump, and the compression of air in a diesel engine.

5. How is the adiabatic process used in practical applications?

The adiabatic process is used in various practical applications, such as in refrigeration and air conditioning systems, where the compression and expansion of gases is used to transfer heat. It is also used in the study of atmospheric processes, such as the formation of clouds and thunderstorms.

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