A lake at an elevation of 15,000 feet, where the barometric pressure is 515.7 mm

Question

A lake at an elevation of 15,000 feet, where the barometric pressure is 515.7 mm Hg.
-What is the barometric pressure in units of atm?
-What is the partial pressure of oxygen at this elevation?
-What is the maximum concentration of oxygen in water (in mg/L) under these conditions, at a temperature of 25.0°C? A lake at an elevation of 15,000 feet, where the barometric pressure is 515.7 mm Hg.
-What is the barometric pressure in units of atm?
-What is the partial pressure of oxygen at this elevation?
-What is the maximum concentration of oxygen in water (in mg/L) under these conditions, at a temperature of 25.0°C? A lake at an elevation of 15,000 feet, where the barometric pressure is 515.7 mm Hg.
-What is the barometric pressure in units of atm?
-What is the partial pressure of oxygen at this elevation?
-What is the maximum concentration of oxygen in water (in mg/L) under these conditions, at a temperature of 25.0°C?

Explanation / Answer

a)

P in atm...

1atm = 760 mm Hg

then

x atm = 515.7 mm Hg

x = 515.7 /760 = 0.6785 atm

b)

Partial pressure if x-O2 = 0.21 approx

P-O2 = x-O2 * Ptotal = 0.21*0.6785 = 0.142485 atm of O2

c)

max O2 in water under this

Recall that Henry's Law models solubility of GASES in liquids. Typically, as Pressure in the gas phase increases, the solubility of the gas increases. As Temperature increases, the solubility of gas will decrease ( contrary to salts)

Note that, for Henry's law to be applicable, the solute must be very low, i.e. << 0.1 (mol fraction)

S = Ki*Pi

Solubility of Gas in mol per liter = S

Hi = Henrys constant

Pi = Partial pressure of

K-O2 = 1.3*10^-3

S-O2 = K-O2 * P-O2 = (1.3*10^-3)*0.142485 = 0.0001852 M o fO2 / Liter

change to mass

mol = 32 g

0.0001852 mol --> 32*0.0001852 = 0.0059264 g / L = 0.0059264*10^3 mg /L = 5.9264 mg of O2 /L

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