Problem 9.10 At high altitudes, mountain climbers are unable to absorb a suffici

Question

Problem 9.10

At high altitudes, mountain climbers are unable to absorb a sufficient amount of O2 into their bloodstreams to maintain a high activity level. At a pressure of 1 bar, blood is typically 95%saturated with O2, but near 18,000 feet where the pressure is 0.50 bar, the corresponding degree of saturation is 71%. Air contains 20.99(%\) O2 by volume. Assume that the density of blood is 998 kg m?3.

Part A

Assuming that the Henry's law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for pressure of 1 bar.

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Part B

Assuming that the Henry's law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for pressure of 0.500 bar.

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Problem 9.10

At high altitudes, mountain climbers are unable to absorb a sufficient amount of O2 into their bloodstreams to maintain a high activity level. At a pressure of 1 bar, blood is typically 95%saturated with O2, but near 18,000 feet where the pressure is 0.50 bar, the corresponding degree of saturation is 71%. Air contains 20.99(%\) O2 by volume. Assume that the density of blood is 998 kg m?3.

Part A

Assuming that the Henry's law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for pressure of 1 bar.

Express your answer with the appropriate units.

mO2 =

SubmitMy AnswersGive Up

Part B

Assuming that the Henry's law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for pressure of 0.500 bar.

Express your answer with the appropriate units.

mO2 =

SubmitMy AnswersGive Up

Explanation / Answer

By Daltons law, Po2 = 0.2099 P

n = nH2O*xO2 = nH2O * Po2/KhO2 = density of H2O * V * Po2 * f/ (KhO2 * MH2O)

= 998 * 10^-3 * 0.2099 * 0.95/(4.95 * 10^4 * 18.02 * 10^-3) = 2.31*10^-4 gm

Mass of O2 = nM = 2.31*1.^-4 * 32.0 = 7.14*10^-3 gm

At 18000ft,

n = 998*10^-3*0.2099*0.50*0.71/(4.95*18.02*10^-3) = 8.34*10^-5 mol

Mass of O2 = nM = 8.34*10^-5*32.0 = 2.67*10^-3 gm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.