A gas has a solubility of 1.46 g L at 8.00 atm of pressure. What is the pressure of a sample of the same gas that
contains 2.7 g/L of the dissolved gas?

Respuesta :

Answer:

[tex]14.8\ \text{atm}[/tex]

Explanation:

[tex]C_1[/tex] = Initial concentration = 1.46 g/L

[tex]C_2[/tex] = Final concentration = 2.7 g/L

[tex]P_1[/tex] = Initial pressure = 8 atm

[tex]P_2[/tex] = Final pressure

From Henry's law we have the relation

[tex]\dfrac{C_2}{C_1}=\dfrac{P_2}{P_1}\\\Rightarrow P_2=\dfrac{C_2}{C_1}P_1\\\Rightarrow P_2=\dfrac{2.7}{1.46}\times 8\\\Rightarrow P_2=14.8\ \text{atm}[/tex]

The pressure of a sample of the same gas at the required concentration is [tex]14.8\ \text{atm}[/tex].