Whats the difference between q and q_v in thermodynamics?

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1. Mar 5, 2015

x86

In thermodynamics,

(delta internal energy) = (delta) U = q + w

But it is also known that (delta U) = q_v = integral of (C_v) dT

So if C_v is constant, then we have (delta U) = q_v = C_v (delta T)

But the confusing part for me is what follows. For instance, in an adiabatic process, for (delta) U = q + w, it is known that q = 0. So (delta) U = w

But then again, people will often use (delta U) = q_v = C_v (delta T) for adiabatic processes.

This is confusing to me,

Isn't q_v esentially the same as q? Except q_v is q at constant volume. So shouldn't q_v = 0 when q = 0?

Last edited: Mar 5, 2015
2. Mar 6, 2015

rude man

First, assume a quasi-static process.
Then this is true in general if the process is isochoric (constant volume). (I guess that's what you mean by "q_v". I would avoid that particular nomenclature).
It is also true for any process on an ideal gas (does not have to be isochoric. So you can use q instead of q_v.)
See above. For adiabatic process, dU = dW = -pdV. W here is work done ON the system (not conventional for physicists, but is for chemists, broadly speaking).
Yes, but again I see no need to invoke "q_v". I would prefer CV = {δQ/dT}V.
Also, reserve lower case for molar parameters, e.g. Cv = ncv etc. where n = no. of moles.

3. Mar 6, 2015

x86

But how can we say that the change in internal energy is equal to the heat flow and the work done, and then go on to say its also equal to only the heat flow (Cv*T)? What of the work?

4. Mar 7, 2015

rude man

In an isochoric process there is no work done. (Why?).

5. Mar 7, 2015

x86

Ah, *facepalm*

Because no work is done, since work is defined as w = -P (delta) V, (delta) V = 0

6. Mar 7, 2015

rude man

In an isochoric process there is no work done. (Why?).
EDIT: it is not correct to say that δq = CvdT unless the process is isochoric.

δQ = dU + pdV
In an isochoric process, dV = 0 so
δQ = dU
and δQ = CvdT = dU
so that Cv ={ ∂U/∂T}V. Note that this equation states constant volume.
However, for an ideal gas we have
U = U(T) only, so that
{∂U/∂T}v = dU/dT
So then, for an ideal gas only, we can conclude that
dU = CvdT.
In other words, the requirement for constant volume in the expression for dU drops out if the gas is ideal.
The correct expression for an ideal gas for δQ is
δQ = CvdT + pdV.

7. Mar 7, 2015

rude man

Right, but read my post #6 anyway.