# If the solubility of a gas is 0.5*g*L^-1 at "1 bar" gas pressure, what will be its solubility at "3 bar" gas pressure?

Feb 2, 2017

Approx. $1.5 \cdot g \cdot {L}^{-} 1$...........

#### Explanation:

$\text{Henry's Law}$ relates to the solubility of a gas in a liquid. At a given temperature, the solubility of the gas is proportional to the partial pressure of the gas above the liquid.

i.e. $\text{Gas solubility}$ $\propto$ ${P}_{\text{gas}}$

Or, $\text{Gas solubility}$ $=$ ${k}_{\text{P"P_"gas}}$, where ${k}_{\text{P}}$ is the socalled $\text{Henry's Law gas constant}$.

And thus, ${P}_{\text{initial solubility"=kxx1*"bar}} = 0.5 \cdot g \cdot {L}^{-} 1$.

${P}_{\text{solubility at elevated pressure"=kxx3*"bar}} = 1.5 \cdot g \cdot {L}^{-} 1$.