# On the derivation of mayers relation

We write,

$$U=q+W$$ [first law]

for constant pressure case

$$C_p \Delta T = \Delta q+ nR \Delta T$$

Now I do the same process but keep the evolume constant then

$$U=C_V \Delta T= \Delta q$$

Now I put that in the original equation,

$$C_p - C_v = nR$$

The doubt I have in this derivation is that couldn't the work change in constant volume process due to energy from $$Vdp$$?

And also know how did we know that that $$\Delta q$$ is exactly $$C_v \Delta T$$?

There is no work in a constant volume. Draw a $$PV$$ diagram for constant volume case. As pressure grows there is no volume change, there is no area under $$PV$$ curve. It is similar to heating the metal container. Container "keeps" volume constant(until it blows up).

There are a lot of mistakes in my original post. I have re-done the derivation correcting my mistakes.

Enthalpy is defined as:

$$\Delta H = \Delta U + \Delta PV$$

For a constant pressure process,

$$\Delta H = nC_p \Delta T$$

$$\Delta U = nC_v \Delta T$$

$$\Delta (PV) = P \Delta V= nR \Delta T$$

Hence,

$$nC_p \Delta T = nC_v \Delta T + nR \Delta T$$

$$C_p - C_v = R$$

This basically relates to the energy change co-efficient of constant pressure and constant volume process. Also, there is no need for $$Vdp$$ work here as we had assumed an isobaric process from the start.

Reference