Question

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

Answer 1

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

`PV = nRT`

`D = (nM)/V = (PM)/(RT)`

where `D = (nM)/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 `"0.804 g/L"` at `"1 atm"`. We can assume that it doesn't change in `"50 K"` going from `100^@ "C"` to `150^@ "C"`. Keep this density in mind. The actual density of gaseous `CH_4` is about `"0.656 g/L"` at `"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.

`barV_(H_2O) = "L"/("0.804 g") xx ("18.015 g")/("1 mol") = "22.407 L/mol"`

`barV_(CH_4) = "L"/("0.654 g") xx ("16.0426 g")/("1 mol") = "24.530 L/mol"`

The molar volume at `150^@ "C"` would be:

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

`=` `"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 `"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.