On the derivation of mayers relation

Question

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 $C_v Delta T$?

Constantin · Accepted Answer

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).

Correct answer by Constantin on July 17, 2021

Buraian · Answer

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.

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On the derivation of mayers relation

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