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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 $C_v \Delta T$?

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2 Answers 2

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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).

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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

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