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According to equipartition of energy, the energy ossociated with each degree of freedom is $\frac{K_{b}T}{2}$ for one molecule .

For 'x' molecule which has degree of freedom f it's energy is given by

$U= \frac{f k_{b}Tx}{2} = \frac{f k_{b}RTn}{2}$ , where n is no. of moles

for small change in temperature

$\frac{dU}{dT} = \frac{fnR}{2}$

$C_{v} =\frac{dU}{ndT} = \frac{fR}{2}$ this much given in text book but why we can not fill dU/dT in $C_{p}$ from which we get $C_{p} = \frac{fR}{2}$.

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  • $\begingroup$ From Wikipedia (en.wikipedia.org/wiki/Heat_capacity#Measurement): "Measurements under constant pressure produce larger values than those at constant volume because the constant pressure values also include heat energy that is used to do work to expand the substance against the constant pressure as its temperature increases." $\endgroup$ – probably_someone Apr 5 at 14:53
  • $\begingroup$ i know it but i am not asking about that i am asking why we dU/dT value in Cv and why not in Cp\ $\endgroup$ – rohit143 Apr 5 at 15:10
  • $\begingroup$ Because $c_p=\frac{\partial H}{\partial T}$. For a derivation, see the same Wikipedia article. $\endgroup$ – probably_someone Apr 5 at 15:10
  • $\begingroup$ thanks i got it . $\endgroup$ – rohit143 Apr 5 at 15:25
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Because specific heat at constant pressure is defined in terms of enthalpy $h$ and not internal energy $u$. It is defined as

$$C_{P}=\biggl(\frac {δh}{δT}\biggr)_P$$

PROOF:

Since

$$h=u+Pv$$

$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δu}{δT}\biggr)_{P}+\biggl(\frac{δ(Pv)}{δT}\biggr)_{P}$$

$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δQ}{δT}\biggr)_{P}-\biggl(\frac{Pδv}{δT}\biggr)_{P}+\biggl(\frac{Pδv}{δT}\biggr)_{P}+\biggl(\frac{vδP}{δT}\biggr)_{P}$$

The last term is zero, therefore

$$\biggl(\frac{δh}{δT}\biggr)_{P}=\biggl(\frac{δQ}{δT}\biggr)_{P}=C_{P}$$

Hope this helps.

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