Question

2. A gas has a solubility in water
at 0°C of 3.6 g/L at a pressure
of 1.0 atm. What pressure is
needed to produce an aqueous
solution containing 9.5 g/L of
the same gas at 0°C?​

Answer 1

Answer:

2.6 atm

Explanation:

At constant temperature, solubility of gas increases as pressure increases, Hence, they varies directly proportional.

i.e  S ∝ P

$\frac{S}{P}= K$

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

where:

S₁ and P₁  are the initial solubility and pressure of the gas

S₂ and P₂ are the final solubility and pressure of the gas

Making P₂ the subject of the formula from the above equation; we have:

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

where; it is given from the question that:

P₁  = 1.0 atm

S₁ = 0.36 g/L

S₂ = 9.5 g/L

Replacing the values into the above equation; we have:

$P_2 = \frac{1.0 atm *9.5 g/L}{0.36 g/L}$

P₂ = 2.6 atm

∴  The pressure needed to produce an aqueous  solution containing 9.5 g/L of  the same gas at 0°C = 2.6 atm