Thermodynamics: Relationship between deltaX, partialX, dx

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1. Jan 28, 2016

ybhathena

1. The problem statement, all variables and given/known data

I am trying to understand the the following derivation:

$Cv = (qv/ΔT) = (ΔU/ΔT) \\ Cv = (∂U/∂T)v \\ dU = CvdT$

3. The attempt at a solution

So here is what I understand so far. I understand that heat transfer q and temperature T are related by a direct proportionality constant C. I also understand that change in internal energy is equal to heat transfer. What I don't understand is how the book goes from change in deltaU and deltaT to partial derivative of U with respect to T while volume is constant (i.e. how did they go from line 1 to line 2).

(For small deltaT U is a linear function of T, deltaU = CvdeltaT)

I phrased my question as "Relationship between deltax, partialX and dx" because I suspect my confusion lies in not understanding what the relationship is between these different symbols.

Thank you very much

Last edited: Jan 28, 2016
2. Jan 28, 2016

haruspex

The derivative is just the limit as the deltas tend to zero. Since we have Cv on the left, we are necessarily discussing change in temperature with volume constant (no?).
Would you be more confortable if it said Cv=(qv/ΔT)|v=(ΔU/ΔT)|v?

3. Jan 28, 2016

Staff: Mentor

For single phase pure materials in general, internal energy U is a function of both temperature T and specific volume V. So U = U(T,V). But, for incompressible solids and liquids, U is a function only of T, so U = U(T). And for ideal gases, internal energy is insensitive to specific volume, so here again, U=U(T). But, for an ideal gas, if the volume of the gas changes (i.e., the gas does work), q is not equal to ΔU; yet ΔU still equals $C_vΔT$. So, in the case of an ideal gas, the only reason it is still called $C_v$ is that, if we want to measure Cv using the amount of heat transferred q, we need to do it at constant volume.

In freshman physics we were taught that C is defined by q = CΔT. But now, in thermodynamics, we learn that q is path dependent (i.e., varies with different process paths between the same two equilibrium states of a system), while C is supposed to be a physical property of the material that depends only on state and not path. So we conclude that a more advanced definition of heat capacity is required. That definition, which is still consistent with what we learned in freshman physics for solids and liquids, is
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
So, in general,
$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$
For ideal gases, and for incompressible solids and liquids, the volume-dependent terms are zero.

4. Jan 29, 2016

ybhathena

But then what is the point of invoking partial derivatives.

5. Jan 29, 2016

haruspex

Since V is taken as constant, and a varying V would affect U, it is more exact to write it as a partial derivative. More generally, V and T might both vary, perhaps according to some functional relationship. In that case, dU/dT would have to take into account the effect of the varying V, but the partial derivative does not.

6. Jan 30, 2016

ybhathena

I'm sorry, I still don't understand this part: Since V is taken as constant, and a varying V would affect U, it is more exact to write it as a partial derivative.

7. Jan 30, 2016

haruspex

In general, U is a function of both V and T. The equation is assuming V is held constant, so it is more accurate to write it as a partial derivative with respect to T. You could write it as dU/dT, provided you add the rider that V is to be held constant. Writing it as a partial avoids the need to mention that.

8. Jan 30, 2016

ybhathena

That makes sense thank you! But why did we have to rewrite the deltas as partial derivatives? You said this earlier: The derivative is just the limit as the deltas tend to zero. Does that mean that partialderivative U/ partialderivative T is the same as saying the value difference between the two internal energies becomes very small divided by the the value difference between the two temperatures becomes very small?

9. Jan 30, 2016

haruspex

Yes, assuming you mean "the limit of (the value difference between the two internal energies) divided by (the value difference between the two temperatures), as the difference in temperatures becomes very small"