Calculate the solubility of dinitrogen in blood that is in contact with air (78% N2) at atmospheric pressure.?
1. Calculate the solubility of dinitrogen in blood that is in contact with air (78% N2) at atmospheric pressure.
_____M
2. Calculate the solubility of dinitrogen in blood that is in contact with air (78% N2) at 4.00 atm, the pressure at a sea depth of 100 feet.
_____ M
3.Determine the volume of N2 that will be released from one liter of blood, if a deep-sea diver surfaces quickly from a sea depth of 100 feet.
_____ L
4. If this release of N2 occurs in the form of gas bubbles that are 1.3 mm in diameter, how many bubbles per liter is this?
_____bubbles
Answer:
You forgot to mention the units of the Henry constant. But i think it is mol/(Latm). The value given is close to solubility of nitrogen in pure water at 298K
For this units the solubility equilibrium is given by
c(N₂) = k · p(N₂)
p(N₂) is the partial pressure it is:
p(N₂) = 0.78 · p(air)
1.
c(N₂) = 3.8·10⁻⁴mol/(Latm) · 0.78 · 1atm = 2.964·10⁻⁴M
2.
c(N₂) = 3.8·10⁻⁴mol/(Latm) · 0.78 · 4atm = 1.1856·10⁻³M
3.
The pressure difference between sea level and depth h is
Δp = ρ_water · g · h
= 1000kg/m³ · 9.81m/s² · 30.48m = 299009Pa = 2.95atm
The difference in solubility is
Δc = k · 0.78 · Δp
= 3.8·10⁻⁴mol/(Latm) · 0.78 · 2.95atm
= 8.747·10⁻⁴mol/L
One liter of water releases n=8.747·10⁻⁴mol nitrogen.
At standard conditions on the surface it represents a volume of
V = N · R · T / p
= 8.747·10⁻⁴mol · 8.314472J/(molK) · 298.15K / 101325Pa
= 2.14·10⁻⁵m³
= 0.0214L
4.
The number of bubbles is total Volume divided by bubble volume:
N = V / ( π·d³/6)
= 2.14·10⁻⁵m³ / ( π · (0.0013m)³/6)
= 18602
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