Question

At NTP, the solubility of natural gas in water is 0.8 mole of gas/kg of water. What is the Henry’s law constant for natural gas?

• A. $8kN/m^2$
• B. $7.90\times {10}^{-3}Pa$
• C. 71.36 bar
• D. 105 mmHg

Answer 1

The correct option is C 71.36 bar

Given,
Solubility of natural gas, S = 0.8 mole of gas/kg of water = 0.8 molal
Hence, moles of natural gas in mixture, $n_{NG}=0.8\text{mole}$
At NTP, pressure, $P_{NG}=1.01325bar$
Number of moles of water in 1000 g, $n_{\text{water}}=\frac{\text{mass}}{\text{molar mass}}$
$n_{\text{water}}=\frac{\text{1000 g}}{\text{18~g/mole}}=55.56mol$

Mole fraction of natural gas in the mixture, $X_{NG}=\frac{n_{NG}}{n_{NG}+n_{\text{water}}}$
$X_{NG}=\frac{0.8}{0.8+55.56}=0.0142$
Using Henry’s law $P_{NG}=K_H\times X_{NG}$
$K_H=\frac{1.01325bar}{0.0142}=71.36bar$.