Question

The solubility of gas is 0.58 g/L at a pressure of 104 kPa. What is the solubility if it’s solubility pressure increases to 250 kPa at the same temperature?

Answer 1

Answer : The solubility of gas at pressure 250 kPa is, 1.39 g/L

Explanation:

According top the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

$S\propto P$

or,

$\frac{S_1}{S_2}=\frac{P_1}{P_2}$

where,

$S_1$ = initial solubility of gas = 0.58 g/L

$S_2$ = final solubility of gas = ?

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 250 kPa

Now put all the given values in the above formula, we get the final solubility of the gas.

$\frac{0.58g/L}{S_2}=\frac{104kPa}{250kPa}$

$S_2=1.39g/L$

Therefore, the solubility of gas at pressure 250 kPa is, 1.39 g/L

Answer 2

Converting to pascal
104/1000=0.104=P1
250/1000=0.25=P2
If 0.58=0.104
=0.25
Cross multiply
Solubility =[0.58*0.25]/0.104=1.39g/l