I need to prove that it is equal to the integral of $C_pdT$ from $T_1$ to $T_2$. But should not be the case only when the pressure is constant?


The enthalpy of an ideal gas does not depend on pressure. It is a function only of temperature. So $C_pdT$ is always equal to dH for an ideal gas of constant composition. The subscript p on $C_p$ merely indicates that $C_p$ can be measured directly by determining the heat Q absorbed by the gas in a constant pressure experiment.


Enthalpy is defined as $$H= E + PV$$

Its change is defined by the differential $$\mathrm dH= \mathrm d E +\mathrm d\, (PV) \;.$$

Using Thermodynamic Identity $$\mathrm d E= - P\mathrm dV + T\mathrm dS $$ for quasistatic process, we get $$\mathrm dE= \delta W + \delta Q$$

So, infinitesimal change in enthalpy becomes $$\begin{align}\mathrm dH &= \delta Q + \delta W +P\;\mathrm d V + V\;\mathrm d P\\ &= \delta Q + V\mathrm d P\;.\end{align}$$

Now, take again the thermodynamic identity

\begin{align}\mathrm d E& = T\;\mathrm dS- P\;\mathrm dV\\ &=\mathrm d(TS)- S\;\mathrm dT -\mathrm d (PV)+ V\;\mathrm dP\\ \implies \mathrm d(E+ PV - TS)&=-S\;\mathrm dT + V\;\mathrm dP\\& = \mathrm dG \;_;\end{align} this implied change in $G$ (which is Gibbs Free Energy) depends on the variation of $T$ and $P\;.$

Therefore we can write; $$\mathrm dG= \left(\frac{\partial G}{\partial T}\right)_P\, \mathrm dT +\left(\frac{\partial G}{\partial P}\right)_T\, \mathrm dP \,.$$ Comparing the above equations, we get $$\left(\frac{\partial G}{\partial T}\right)_P= -S\\\left(\frac{\partial G}{\partial P}\right)_T= V $$

Equating the cross derivatives, we get $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P\;.$$

Also, $$\mathrm d S = \frac{\mathrm dE}{T} + \frac{P}{T}\, \mathrm d V\,$$ and this implies $S$ depends on $T, V\;.$

Now, the definition of $C_P$ is $$C_p = \left(\frac{\partial Q}{\partial T}\right)_P= T\,\left(\frac{\partial S}{\partial T}\right)_P\;.$$

Using all these, we can write the differential of $H$ as

\begin{align}\mathrm dH &= T\left[\left(\frac{\partial S}{\partial T}\right)_V \,\mathrm dT+ \left(\frac{\partial S}{\partial V}\right)_T \,\mathrm dV\right] + V\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial V}{\partial T}\right)_P\right]\,\mathrm dP\\ &=C_P\,\mathrm dT - V\left[1 - \frac{T}{V}\,\left(\frac{\partial \left(\frac{\nu R T}{P}\right)}{\partial T}\right)_P\right]\,\mathrm dP \\ &= C_P\,\mathrm dT - V\left[1 - \frac{\nu R T}{PV}\,\left(\frac{\mathrm d T}{\mathrm dT}\right)\right]\,\mathrm dP\\ &= C_P\,\mathrm dT\;.\end{align}

So, the total change in enthalpy $$\Delta H= \int_{P_1,T_1}^{P_2,T_2}\; C_p\,\mathrm dT\;.$$


$\bullet$ F Reif's Fundamentals of Statistical and Thermal Physics.

$\bullet$ E Fermi's Thermodynamics.

  • $\begingroup$ That equation for $\Delta H$ is incorrect. It should be calculated from $$dH=C_pdT+\left(V-T\frac{\partial V}{\partial T}\right)dP$$For an ideal gas, this reduces to $\Delta H=C_pdT$. $\endgroup$ – Chet Miller Feb 19 '16 at 1:56
  • $\begingroup$ The starting equation for dH should be dH=TdS+VdP. One then writes $$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$. Then, there is a Maxwell relation giving $$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$There is also the equation $$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$If you combine all these equations, you obtain the equation that I wrote. $\endgroup$ – Chet Miller Feb 19 '16 at 2:52
  • $\begingroup$ This is a well-known derivation that is found in most Thermodynamics textbooks. Please just consider presenting the final result with a reference. $\endgroup$ – Chet Miller Feb 19 '16 at 3:13

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