# Which gas at 150^@ "C" is more ideal at "1 atm", water or methane?

Mar 24, 2017

You can look at their densities. The ideal gas law in two forms is:

$P V = n R T$

$D = \frac{n M}{V} = \frac{P M}{R T}$

where $D = \frac{n M}{V}$ is the density, $P$ is the pressure, $V$ is volume, $M$ is molar mass, $n$ is mols, $R$ is the universal gas constant, and $T$ is temperature.

The actual density of gaseous water is $\text{0.804 g/L}$ at $\text{1 atm}$. We can assume that it doesn't change in $\text{50 K}$ going from ${100}^{\circ} \text{C}$ to ${150}^{\circ} \text{C}$. Keep this density in mind. The actual density of gaseous $C {H}_{4}$ is about $\text{0.656 g/L}$ at $\text{1 atm}$. Keep this in mind.

The density is inversely proportional to the molar volume; simply reciprocate the density and multiply by the molar mass.

${\overline{V}}_{{H}_{2} O} = \text{L"/("0.804 g") xx ("18.015 g")/("1 mol") = "22.407 L/mol}$

${\overline{V}}_{C {H}_{4}} = \text{L"/("0.654 g") xx ("16.0426 g")/("1 mol") = "24.530 L/mol}$

The molar volume at ${150}^{\circ} \text{C}$ would be:

barV = (RT)/P = (("0.082057 L"cdot"atm/mol"cdot"K")("423.15 K"))/("1 atm")

$=$ $\text{34.722 L/mol}$

The ideal gas law overestimates the molar volume and thus underestimates the density. Since the ideal gas law was less far off for methane, methane is more ideal at $\text{1 atm}$ than water is.

This makes sense because one might expect some weak long-range dipole-dipole interactions (if any) between water molecules in the gas phase, whereas methane definitely has none of that.