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I am supposed to derive an expression for the specific heat capacity $c_p$ for air with a water vapour mass mixing ratio of $\mu$. The hint is: (Use the fact that $c_p= (\frac{\partial H}{\partial T})_p$, where H is the enthalpy per unit mass.

I do not want a complete solution, but I have absolutely no idea where to start. What is the relation between enthalpy and mixing ratio, enthalpy is just internal energy and isobaric work.

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  • $\begingroup$ Actually, enthalpy (per mole) is just equal to U + PV. For an ideal gas mixture like air and water vapor, PV = RT. So, for an ideal gas H = U + RT. Mixing ratio is just another name for mole fraction. In an ideal gas mixture, the heat capacity of the mixture is equal to the heat capacities of the individual pure components, weighted in terms of their mole fractions. $\endgroup$ – Chet Miller Feb 6 '16 at 11:49
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Enthalpy of a mixture of gases is determined by this formula: $$h=\Sigma (mf)_ih_i\;[\mathrm {kJ/kg}]$$ $h$ is the enthalpy of unit mass of the mixture,

$(mf)_i$ is mass fraction of gas $i$, $(mf)_i=\large{\frac{m_i}m}$ ($m_i$ is the mass of gas $i$ and $m$ is the mass of mixture)

$h_i$ is the enthalpy of unit mass of gas $i$ that can be determined by using $\textrm{one}^1$ of Dalton's law of additive pressures or Amagat's law of additive volumes and temperature of the mixture.

In this case, if we assume that $\large{\frac{m_v}{m_a}}=\mu$; as $m=m_v+m_a$, then we have $$(mf)_v=\large{\frac{\mu}{\mu+1}}$$ and $$(mf)_a=\large{\frac 1{\mu+1}}$$ Thus, $$\large{h}=\large{\frac{\mu}{\mu+1}}h_v+\large{\frac 1{\mu+1}}h_a$$


$^{\mathrm 1}$ For mixture of ideal gases, both of these laws end to same result, but for mixture of real gases, choosing between these two laws depends on the problem.

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