# Confusion of the Isobraic Process Heat equation

I was currently reading a wiki article on Isobaric Process (Link at bottom f post). And am confused as to why in there derivation of heat of a an isobaric system they use specific heat capacity for volume when volume is changing and pressure is constant, so why they relating internal energy to a heat capacity of a constant volume and not pressure and how is $$C_p=C_v+R$$, as in my text book 'Physics for scientist and engineers' by tippler it states that $$C_p-C_v=nR$$, so where has the n gone from the wiki equation?

$$\underline{Screen \: Capture \: Of \: Wiki}$$ https://en.wikipedia.org/wiki/Isobaric_process

• This question frequently arises; see here for discussion and links to other answers. – Chemomechanics Mar 23 '19 at 20:36

The reason why the specific heat at constant volume is used even though it is not a constant volume process is because it happens to be that change in internal energy for an ideal gas, for ANY process, is given by

$$\Delta U = C_v\Delta T$$

This can be derived as follows:

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 $$\Delta U=C_p\Delta T – R\Delta T$$

Also, for any ideal gas

$$R=C_p-C_v$$

Substituting

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

Concerning the equation for R, $$C_p-C_v=R$$ for one mole of an ideal gas (n=1)

We can prove that $$C_p-C_v=R$$ is by using the basic 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 $$C_p – C_v = R$$

Hope this helps.