I'm trying to get nitrogen dissolved into my homebrewed beer. I have a nitrogen tank that I'm going to hook to a keg that's rated for 130 PSI (with a PRV rated at 100 PSI).

Is there any way to reliably calculate the solubility of nitrogen in water, or the amount of dissolved nitrogen in water, in high pressures and low temperatures (up to 80 PSI an -1c)? Does Henry's Law apply here?

  • 1
    $\begingroup$ 130 PSI is not a high pressure, at least not in context of physical chemistry. Solubility of nitrogen can be safely assumed to be proportional to pressure in that range. // For carbon dioxide, it would be more complicated due high solubility and several chemical equilibrii. $\endgroup$
    – Poutnik
    Jan 5 at 12:25
  • $\begingroup$ If tempted to dramatically lower the temperature of the keg, please check the MDMT (minimum design metal temperature) if listed. If not listed, do not go below freezing, if that. $\endgroup$
    – Jon Custer
    Jan 5 at 14:19

1 Answer 1


I will propose you one scheme for estimating its value, with the assumptions I think are reasonable. The data employed are found at the end.

The gas phase is assumed to be pure $\ce{N2}$ and we will consider its deviations with ideality, even though the pressure is not that high ($\approx \pu{5.5 atm}$) and nitrogen is a non-polar molecule. The liquid phase is assumed to be a mixture of $\ce{N2 + H2O}$. Since $\ce{N2}$ is barely soluble in water, we will model the non-idealities with Henry's law.

The chemical equilibrium, and thus the molar fraction of nitrogen in the liquid phase turns out to be

$$ \phi_\ce{N2}P = H_\ce{N2,H2O}x_\ce{N2} \rightarrow x_\ce{N2} = \frac{\phi_\ce{N2}P}{H_\ce{N2,H2O}}\tag{1} $$

Evaluating Eq. (1) anwers the question, but we need to estimate two factors: (1) the fugacity coefficient of nitrogen in the gas phase $\phi_\ce{N2}$, and (2) the Henry's constant for nitrogen in water as solvent $H_\ce{N2,H2O}$.

  • $\phi_\ce{N2}$: I will model the gas phase with the virial equation of state up to the second virial coefficient. In this case $$ \phi_\ce{N2} = \exp\left(\frac{BP}{RT}\right) \tag{2} $$ where the second virial coefficient $B$ is calculated with statistical thermodynamics, using the Lennard-Jones potential for the intermolecular potential function \begin{align} \Gamma(r) &= 4 \varepsilon\left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right] \tag{3} \\ B &= 2 \pi N_\mathrm{A} \int_0^\infty \left\{1 - \exp\left[-\frac{\Gamma(r)}{kT}\right]\right\} r^2 \mathrm{d}r \tag{4} \end{align} Combining Eqs. (3) and (4), we can then numerically integrate to obtain $B$. Then, we go back and calculate Eq. (2).
  • $H_\ce{N2,H2O}$. This constant is a function of temperature, but only of pressure at sufficiently high values. One important assumption is that I will neglect the dependence of $H_\ce{N2,H2O}$ on pressure, and only take into account the temperature. An empirical equation for this is given in Benson and Krause (1976) \begin{equation} \ln\left(\frac{H_\ce{N2,H2O}}{\pu{bar}}\right) = \alpha_2 \left(1 - \frac{T_2}{T}\right) - \beta \left(1 - \frac{T_2}{T}\right)^2 + \ln(1.01325) \tag{5} \end{equation} with $\alpha_2$, $T_2$, $\beta$ being fitting parameters.

Fixing the pressure to $\pu{130 psi}$, we can repeat the calculations for many temperatures. The results are shown below:

enter image description here

The minimum in the solubility is due to the dependency of Henry's constant with temperature. When its value is the highest, the tendency of $\ce{N2}$ to leave the liquid phase is the highest, and thus the molar fraction acquires a minimum value. The estimation you want is when the red curve hits the left $y$ axis, which is \begin{equation} \boxed{x_\ce{N2} \approx \pu{1.3 \cdot 10^{-4}}} \end{equation}

The parameters for the intermolecular potential function for nitrogen are: \begin{equation} \sigma = \pu{3.667*10^{-10} m} \hspace{1 cm} \varepsilon = \pu{1.378 \cdot 10^{-21} J}; \end{equation}

The constants for the correlation of the Henry's constant are: \begin{equation} T_2 = \pu{162.02 K} \hspace{1 cm} \alpha_2 = 41.712 \hspace{1 cm} \beta= 36.855; \end{equation}


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