(1)What is atmospheric pressure on top of Mt. Everest on a day when water boils

Question

(1)What is atmospheric pressure on top of Mt. Everest on a day when water boils there at a temperature of 70.0°C?
_____N/m2

(2)A deep sea diver should breathe a gas mixture that has the same oxygen partial pressure as at sea level, where dry air contains 20.9% oxygen and has a total pressure of 1.01 105 N/m2.

(a)What is the partial pressure (in N/m2) of oxygen at sea level?

_____N/m2

(b)If the diver breathes a gas mixture at a pressure of 1.00 106 N/m2, what percent oxygen should it be to have the same oxygen partial pressure as at sea level?

Explanation / Answer

(1)What is atmospheric pressure on top of Mt. Everest on a day when water boils there at a temperature of 70.0°C?
if T boils at T = 70 C

apply Classisus equation

ln(P2/P1) = H/R*(1/T1-1/T2)

ln(P2/1atm) = 40660/8.314*(1/373 - 1/(70+273))

P2 = exp(-1.1467)

P2= 0.31768 atm

1 atm = 101325 Pa = 101325 N/m2

0.31768 atm = 0.31768*101325 N/m2 = 32,188.926 N/m2

(2)A deep sea diver should breathe a gas mixture that has the same oxygen partial pressure as at sea level, where dry air contains 20.9% oxygen and has a total pressure of 1.01 105 N/m2.

(a)What is the partial pressure (in N/m2) of oxygen at sea level?

Recall that:

P-O2 = x-O2 * Ptotal = 0.209*(1.01*10^5) = 21,109 N/m2

(b)If the diver breathes a gas mixture at a pressure of 1.00 106 N/m2, what percent oxygen should it be to have the same oxygen partial pressure as at sea level?

Then...

P-O2 = 21109 N/m2

P-O2 = x-O2 * Ptotal

21109 = x-O2 * 10^6

x-O2 = 21109 / (10^6) = 0.021109

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