16/07/2021
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Hello Students.
Welcome back to NUCLEOCHEM FORUM, A wizard of Chemistry.
Here in this video we will learn about a Henry’s law and its example/s
In this video; we will learn Henrys’ law, solubility of gas, relationship between Henry’s law constant, Solubility and Pressure on the gas.
Details of Henry’s Law:
As gases are compressible; their solubility in the liquid is increases by external pressure of the gas. The solubility of the gases increases with the increase of the pressure.
Henry’s law states that, “The solubility of the gas in liquid at a constant temperature is always directly proportional to the pressure of the gas above the solution”.
Consider ‘S’ is the solubility of the gas in Moles/dm3 at a pressure ‘P’ at constant temperature, then according to Henry’s law:
S ∝ P
∴ S = K . P
Where,
K is known as Henry’s law constant.
If P = 1 atm. then
S = K.
If several gases are present, then the solubility of any gas is directly proportional to its partial pressure at a given temperature.
Illustration of Henry’s law:
In case of any aerated drinks (cold drinks), the bottle is fitted by dissolving CO2 gas at high pressure and then sealed. Above the liquid surface there is air and undissolved CO2. Due to high pressure, the amount of dissolved CO2 is large. When the cap of the bottle is removed, the pressure on the solution is lowered and hence excess of CO2 and air escapes out in the form of effervescence (foam). Thus by decreasing the pressure, solubility of CO2 is decreased.
Units of Henry’s law constant:
According to Henry’s law;
S = K ×P
Where;
S = Solubility of the gas in Moles/dm3
K = Henry’s law constant
P = Pressure of the gas in atm.
Therefore;
S = K × P
K = S/P
K = (Moles/dm3)/atm
∴ K = Moles dm-3 atm-1
Link for Other videos are also given here:
SOLUTION & COLLIGATIVE PROPERTY || SESSION - 1 :
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