# Can ideal gas law be applied to water vapour? [closed]

How to accurately estimate the molar concentration of pure water vapour at 298 K? Can ideal gas law be applied to water vapour to calculate the molar concentration?

• Good or bad, you don't really have any other approximation, do you? Nov 1, 2018 at 17:36
• Get yourself a generalized corresponding states plot (from the internet) of the compressibility factor z as a function of the reduced temperature and the reduced pressure. Knowing the critical pressure and critical temperature of the gas, as well as the specified temperature and pressure, you can judge from the plot (i.e., from the value of z) how accurate the ideal gas law answer will be. For the state that you have defined, @Loong's comparison shows that z will be about 0.998. Nov 2, 2018 at 2:47
• It depends on the range of pressure you want to apply the ideal gas law to, and your definition of accuracy. Nov 2, 2018 at 10:35

Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour.

However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $$T=298\ \mathrm K$$ alone is not enough. The ideal gas law does not give you a value for the second quantity.

For example, assuming equilibrium conditions for saturated steam at a temperature of $$T=298.00\ \mathrm K$$, the corresponding pressure is $$p=3141.7\ \mathrm{Pa}$$.

Using the ideal gas law yields a concentration of

$$\frac nV=\frac p{RT}=\frac{3141.7\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298.00\ \mathrm K}=1.2680\ \mathrm{mol\ m^{-3}}$$

The actual value for water vapour at this temperature and pressure is

$$\frac nV=1.2701\ \mathrm{mol\ m^{-3}}$$

Thus, using the ideal gas law introduces an error of about $$0.2\ \%$$ in this case, which could be acceptable for many practical purposes.