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ss7ja [257]
3 years ago
15

If the solubility of a gas is 8.4 g/L at 404 kPa pressure, what is the solubility of the gas when the pressure is 101 kPa? Pleas

e show work.
Chemistry
1 answer:
Vlada [557]3 years ago
6 0

Answer:

at P = 101 KPa ⇒ S = 2.1 g/L

Explanation:

Henry's law:

  • <em>C</em> = K*P

∴ K: Henry's constant

∴ <em>C</em> is the concentration ≡ Solubiluty (S)

∴ S = (8.4 g/L)*(Kg/1000 g)*(1000 L/m³) = 8.4 Kg/m³

∴ P = (404 KPa)*(1000 Pa/KPa) = 404000 Pa

∴ Pa = Kg/ms²

⇒ S = K(404000 Pa) = 8.4 Kg/m³

⇒ K = 2.079 E-5 s²/m².......the henry's constant for this gas

at P = 101 KPa = 101000 Pa

⇒ S = (2.079 E-5 s²/m²)*(101000 Kg/ms²)

⇒ S = 2.1 Kg/m³ = 2.1 g/L

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