thermodynamics - Why is $dE = C_V dT$ for an ideal gas, even when the process is not constant volume?

I have come across the following derivation for the "equation of an adiabat for an ideal gas" in many textbooks. The following steps are referenced from a separate question:

1. dE=dQ+dW


2. dW=−pdV


3. dQ=0


4. dE=$C_{V}$dT


5. therefore $C_{V}$dT=−pdV

All the symbols have the usual meaning.(The derivation then proceeds to use the eqn. of state arriving at PV$^{\gamma}$=const.)

Problem

1. I don't understand the use of eqn. 4 in step 5. In an adiabatic process both P and V can vary so, then how can one use a quantity that requires V to be constant?

Answer

For an ideal gas any process

$$\Delta U =C_{v}\Delta T$$

so even though it is not a constant volume process it still applies.

I can give you a proof if you need it.

Here is the proof. Actually it is not a proof, but shows that it is true for the examples of an isobaric and adiabatic process. You can do the same for an isothermal or any other process.

For a constant pressure process:

$$\Delta U=Q-W$$ $$\Delta U=C_p\Delta T – P\Delta V$$ For one mole of an ideal gas $$P\Delta V=R\Delta T$$ Therefore $$\Delta U=C_p\Delta T – R\Delta T$$

For an ideal gas,

$$R=C_p-C_v$$

Therefore,

$$\Delta U=C_p\Delta T – (C_p-C_v)\Delta T$$ $$\Delta U=C_v\Delta T$$

For an adiabatic process (Q=0):

$$\Delta U=-W$$ $$\Delta U=- \frac {R\Delta T}{1-k}$$ For an ideal gas $$k=\frac{C_p}{C_v}$$ and again $$R=C_p-C_v$$ Therefore $$\Delta U=- \frac{(C_p-C_v)\Delta T}{1-C_p/C_v}$$ $$\Delta U= C_V\Delta T$$

So you might ask, what is the proof that for an ideal gas $C_p-C_v=R$. It is based on the definitions of the specific heats and enthalpy, combined with the ideal gas law.

Specific heat definitions, ideal gas (they are actually partial derivatives holding P and V constant, respectively):

$$C_p = \frac {dH}{dT}$$ $$C_v = \frac {dU}{dT}$$ Definition of enthalpy (H) $$H = U + PV$$ For one mole of an ideal gas, ideal gas law $$PV=RT$$ Therefore $$H = U+RT$$

Taking the derivative of the last equation with respect to temperature:

$$\frac {dH}{dT} =\frac {dU}{dT}+R$$ Substituting the specific heat definitions into the last equation, we get $$C_p – C_v = R$$

Finally, as J. Murray points out, this only applies to an ideal gas.

Hope this helps.