The easiest way would be to use statistical physics. Gases are effectively described with a one-particle distribution function:
$$f \sim \exp \left[-\frac{E}{kT} \right]$$
which leads to a spatial part of $f$ (concentration) in a uniform gravity field to be
$$n_i \sim \exp \left[-\frac{m_i g z}{kT} \right]$$
That's the easiest way to show that "the ratio of the heavy constituent to the lighter constituent increases as you decrease in altitude":
$$\frac{n_1}{n_2} \sim \exp \left[\frac{(m_2-m_1) g z}{kT} \right]$$
But you seem to be interested in some sort of thermodynamic derivation. You should understand that for it you need to know the form of energy and entropy of the substance in hand, you can't get these from thermodynamics itself.
For a simplicity assume the gases to be held in a cylinder of height $h$ and area $F$:
$$U = \sum \left[ C_i N_i T + F \int_0^h n_i(z) m_i g z \, dz \right]$$
$$S = \sum_i \left[S_{0i} + C_i N_i \ln T - F \int_0^h k \, n(z) \ln n_i(z) \, dz \right]$$
Thus Helmholtz free energy $U-TS$ is:
$$A = \sum_i \left[C_i N_i (T - T \ln T) - T S_{0i} + F \int_0^h n_i \left( m_i g z + k T \ln n_i \right) \, dz \right]$$
Note, that $T$, $F$ and $h$ are constant, as the overall number of particles $N_i$. Minimizing this functional will lead you to equations of the form:
$$m_i g z + k T \ln n_i = const_i$$
which solve the task.
Another way to do it thermodynamically would is to employ chemical potential:
$$ \mu_i + m_i g = const_i$$
Again one need to know the form of the chemical potential for an ideal gas. I assure you, you'll get the right answer if you actually do it. Anyway the equation above and the minimization of the free energy both are derived from the second law, so they are equivalent in essence.
The heat capacity of moist air follows the simplest prescription you can think of,
$$ c_s = 1.005+1.82 H\,\frac{\rm kJ}{\rm kg\cdot {}^\circ C}$$
where 1.005 is the figure for dry air and $H$ is the number of kilograms of water vapor per kilogram of air.
At 20 Celsius degrees, the saturated vapor pressure for water vapor is 2.3 kPa. Sixty percent of it is 1,380 Pa, about 0.014 of the total pressure. This partial pressure doesn't represent the percentage of the mass. Dry air has 29 g/mole or so, water has 16+1+1=18 g/mole, so the mass ratio is 18/29 times 0.014 = 0.0087. Even when multiplied by 1.82, we get 0.016 or so: we may raise 1.005 to 1.02 at most. The humidity doesn't affect the heat capacity too much.
The actual reason for your wildly different number 1,300 is different and simple: you have confused kilograms and cubic meters.
Just to be sure, the latent heat plays no role here because no phases are being changed (condensed/vaporized) during the heating of a decently moist air. The heat capacity of liquid water plays no role, either: no water is actually liquid (in big chunks) in this problem.
The value 1,300 of something is actually the volumetric specific heat capacity of air, so the right units are $\rm kJ/ m^3\cdot {}^\circ C$. The word "volumetric" means that it's computed per unit volume. Divide it by 1.275 kilograms per cubic meters, the density of air, and you will get 1.02 kJ/kg per degree, in agreement with the result two paragraphs above.
The heat capacity of individual gases is simple to compute when one considers the molar heat capacity or, equivalently (up to multiplication by Avogadro's constant), heat capacity of a single molecule. A spherically symmetric atom would have $3k/2$, a diatomic one has $5k/2$, more general molecules that remember the precise orientation of everything have $3k$ etc. Multiply it by $N_A$ to get the molar heat capacities $3R/2$ etc. This may also be divided by the molar mass to get the specific heat capacity etc. When you consider the appropriate weighted average of the molecules/atoms that make up air, you may calculate the heat capacity of the air, too. Thermodynamics of diluted gases is simple because the interactions are so weak so we're really summing simple properties of individual molecules only.
Best Answer
You have to look at Boltzmann factors:
$$ e^{-\frac{E}{kT}} = e^{-\frac{mgh}{kT}}$$
for $m=18u$ and $m=28u$. Roughly at 300K and uniform $g$. For 1 amu, the scale height is:
$$ h \approx \frac{kT}{mg} = \frac{4\times 10^{-21}J}{1.6\times10^{-26}J/m}=250,000m$$
Divide that by 18 or 28 your get a scale height on the order of 10-14km, which is the scale height of the atmosphere. Hence, you don't see any stratification locally.