How is $\mathrm dU=C_V\,\mathrm dT$ always true? [duplicate]

Question

For ideal gases, in case of an isochoric process it is pretty straight forward:

$\mathrm dU=Q+W$

$\Rightarrow\mathrm dU=C_V\,\mathrm dT+p\,\mathrm dV$

$\Rightarrow\mathrm dU=C_V\,\mathrm dT$

But in case of an isobaric process,

$\mathrm dU=Q+W$

$\Rightarrow\mathrm dU=C_p\,\mathrm dT-p\,\mathrm dV$

I tried deriving it this way but I'm stuck. How do I get $\mathrm dU=C_V\,\mathrm dT$ for an isobaric process?

And I have no clue how to derive it for an adiabatic process either, so how do you get the expression for $\mathrm dU$ for an adiabatic process?

Edit: I have thought of substituting $pdV = RdT$ then plugging in $R=C_p-C_v%$. But I was hesitant. Is $du=C_vdT$ because $R=C_p-C_v$ or the other way around?

Answer 1

$U$ is a state function. That means that no matter which path we choose from the point 1 to point 2, we always have the same $\Delta U$ between these two points.

Now in order to prove that

$$\tag{1} \Delta U = n_m C_v \Delta T; \quad \quad n_m \; \text{is the number of moles} $$ all we need to do is find just one process (one path from starting point 1 to finishing point 2) in which (1) holds. That will be enough to prove that (1) always holds.

Now, consider any starting and finishing point (look at the picture)

We can see that that points 1 and 2 (they can be any two points!) can always be connected via 1-A-2, where 1-A is isothermal process whereas A-2 is isochoric process. In process 1-A there is no change in $U$ because in ideal gas $U$ depends only on temperature, so $\Delta U_{1-A} =0$. On the other hand, we can write $\Delta U_{A-2} =n_m C_v \Delta T$ because A-2 is isochoric process.

Thus, we found one path in which (1) holds. So, using the fact that $U$ is a state function we conclude that (1) holds always.

Answer 2

$dU=C_vdT$ is a generic statement, for ideal gases. I will try to convince you of this by proving this from first principles. This is standard, see Reif for example.

Since there are 2 free parameters(the third is determined from PV=RT), we choose $U=U(T,v)$(this makes the calculation simpler than the other choices). Then, $$dU=\bigg( \frac{\partial U}{\partial T}\bigg)_VdT+\bigg(\frac{\partial U}{\partial V}\bigg)_TdV$$.

Similarly, for $s=S(T,V)$ (the entropy)-$$dS=\bigg( \frac{\partial S}{\partial T}\bigg)_VdT+\bigg(\frac{\partial S}{\partial V}\bigg)_TdV$$

Also, $$dS=\frac{1}{T}(dE+PdV)=\frac{1}{T}\bigg(\bigg(\frac{\partial U}{\partial T}\bigg)_VdT+\bigg(\bigg(\frac{\partial U}{\partial V}\bigg)_T+\frac{RT}{V}\bigg)dV\bigg)$$

The last two equations allow us to read off $$\frac{\partial S}{\partial V}, \frac{\partial S}{\partial T}$$(I will drop the subscripts subsequently).Using $$\frac{\partial^2S}{\partial V\partial T}=\frac{\partial^2S}{\partial T\partial V}$$, we get $$\bigg(\partial U/\partial V\bigg)_T=0$$,

i.e. for a given $T$, the energy depends only on $T$. Both $V$ and $P$ may change, but as long as you have specified a $T$, the energy doesn't. Thus, we have, IN GENERAL, $$dU=\bigg(\frac{\partial U}{\partial T}\bigg)_VdT= \bigg(\frac{dQ+pdV}{\partial T}\bigg)_VdT=\bigg(\frac{\partial Q}{\partial T}\bigg)_VdT=C_vdT$$

Answer 3

For an ideal gas at constant pressure, pdV=RdT. So,$$dU=C_pdT-RdT=(C_p-R)dT=C_vdT$$