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

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?

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.

• Yes, sorry. See revision – Bob D Apr 27 '19 at 0:44
• @lineage Note that this is a peculiarity of the ideal gas model, and is not true in general. – J. Murray Apr 27 '19 at 1:13
• @lineage See my revised answer with the proof. And J. Murray is correct that this only applies for an ideal gas. Hope this helps. – Bob D Apr 27 '19 at 1:20
• @lineage see my revised answer with the proof. Hope it helps. – Bob D Apr 27 '19 at 1:26
• thanks for your detailed effort...i didn't understand (under "For an adiabatic process") ΔU=−RΔT/(1−k)..how did you get the denominator?Also since P isn't constant here shouldn't it be W=PΔV=RΔT-VΔP and not W=PΔV=RΔT? – lineage Apr 27 '19 at 1:52

For an ideal gas

ΔU=$$C_{V}$$ΔT

is always true. Hence step 4's use in step 5 is allowed.

Proof:

Consider any reversible process that takes the ideal gas(n=1) from an initial(P1,V1,T1) to final state(P2,V2,T2).

1. Since the process is reversible we can consider infinitesimal differential changes from (P,V,T) to (P+dP,V+dV,T+dT'') as the system goes from initial to final state.
2. At each of these steps, since U is a state var,calculate dU from an attached isobar and isochor betweeen same endpoints.
1. isobar from (P,V,T) to (P,V+dV,T+dT'):
dQ=$$C_{P}$$dT' where dT'=PdV/R (from PV=RT-this is why its valid only for ideal gas)
dW=PdV

2. isochor from (P,V+dV,T+dT') to (P+dP,V+dV,T+dT'+dT'')
dQ=$$C_{V}$$dT'' where dT''=VdP/R (from PV=RT)
dW=$$0$$

3. total dU=$$C_{P}$$PdV/R -PdV + $$C_{V}$$VdP/R-$$0$$
then using $$C_{P}$$-$$C_{V}$$=R (this derivation also requires PV=RT) dU simplifies to $$C_{V}$$dT

1. Since the process is reversible, we can integrate the differential to get
ΔU=$$C_{V}$$ΔT (for an ideal gas $$C_{V}$$ is a T independent constant)

Just imagine a path made up of tiny zigs and zags in it, with the zigs carried out at constant temperature (but with heat added) and the zags carried out at constant volume (but with an equal amount of heat removed, so that, overall, the zig and zag are adiabatic). For the zigs, you would write $$0=dQ-PdV$$and for the zags, you would write $$C_vdT=-dQ$$So for the combination of a zig and zag, you would have the sum of the two tiny changes, given by $$C_vdT=-PdV$$Note that, in the zig part of the path, because internal energy of an ideal gas only depends on temperature, the change in internal energy in a zig is zero.

As you make the zigs and zags smaller and smaller, you approach a continuous adiabatic path.