Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25
̊c is 1.3 x 10-3 mol/l atm. how do you expect the solubility to change if the temperature were decreased?
2 answers:
A. The solubility of O₂ in water : S = 2.052 . 10⁻⁴ mol/L
B. If the temperature were decreased then the solubility of O₂ will increase
<h3>Further eplanation </h3>In the transfer from the gas phase to the liquid phase diffusion process occurs.
The solubility of the gas itself decreases with increasing temperature
The partial pressure of the gas itself is influenced by the concentration of the gas and temperature
Henry's Law states that the solubility of a gas is proportional to its partial pressure
Can be formulated
S = kH. P.
S = gas solubility, mol / L
kH = Henry constant, mol / L.atm
P = partial gas pressure
The partial gas pressure is proportional to the percentage of the gas volume in the mixture
Henry's law constant = 1.3 x 10⁻³mol / l atm.
P = 120 torr = 0.15789 atm
So
S = 1.3 x 10⁻³ . 0.15789
S = 2.052 . 10⁻⁴ mol/L
If the temperature were decreased then the solubility of O₂ will increased
<h3>Learn more </h3>grams of CO will dissolve in 1.00 l of water
the partial pressure of hydrogen gas
the solubility of a gas
Keywords: Henry's Law, partial pressure, gas solubility
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The given data is as follows.
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Q =
or, Q =
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Q =
121300 kJ =
=
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