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 $C_v \Delta T$?


2 Answers 2


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,

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

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

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


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

$$ C_p - C_v = R$$

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.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.