Whats the difference between q and q_v in thermodynamics?

[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?

 

[rude man]

In thermodynamics,

(delta internal energy) = (delta) U = q + w
But it is also known that (delta U) = q_v = integral of (C_v) dT

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.)

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,

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).

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

Yes, but again I see no need to invoke "q_v". I would prefer C_V = {δQ/dT}_V.
Also, reserve lower case for molar parameters, e.g. C_v = nc_v etc. where n = no. of moles.

 

[x86]

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 C_V = {δQ/dT}_V.
Also, reserve lower case for molar parameters, e.g. C_v = nc_v etc. where n = no. of moles.

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?

 

[rude man]

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?

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

 

[x86]

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

Ah, *facepalm*

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

 

[rude man]

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

Start with the 1st law:
δQ = dU + pdV
In an isochoric process, dV = 0 so
δQ = dU
and δQ = C_vdT = dU
so that C_v ={ ∂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 = C_vdT.
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 = C_vdT + pdV.

 

[rude man]

Ah, *facepalm*

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

Right, but read my post #6 anyway.