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標題:
Chemistry(Gas Law)
發問:
Inspired air (breathed in) is about 21% oxygen, 0.93% argon, and the balance nitrogen. (The level of CO2 is negligible at <0.03%). If you lung capacity is 5.50 L at sea level on a standard day, what is the partial pressure of oxygen in your lung?If you climbed Mr. Rainier you would be at an elevation of... 顯示更多 Inspired air (breathed in) is about 21% oxygen, 0.93% argon, and the balance nitrogen. (The level of CO2 is negligible at <0.03%). If you lung capacity is 5.50 L at sea level on a standard day, what is the partial pressure of oxygen in your lung? If you climbed Mr. Rainier you would be at an elevation of 14000 ft above sea level, and the temperature would be -10.0 degree Celsius. What is the partial pressure of oxygen in your lungs at this elevation?
最佳解答:
Suppose that on a standard day and at the sea level, the atmospheric pressure is 1.01 × 105 Pa and the body temperature (in the lungs) is 310 K (37°C). Then by the ideal gas equation: PV = nRT n = PV/RT = (1.01 × 105 × 5.5 × 10-3)/(8.314 × 310) = 0.216 moles which is the total no. of moles of all gas in air. Then by Avogadro's law: Mole ratio is equal to volume ratio under the same temperature and pressure, we have: No. of moles of oxygen = 0.216 × 0.21 moles Further, by the Dalton's law: Partial pressure is proportional to the mole fraction, we have: Partial pressure of oxygen = [(0.216 × 0.21)/0.216] × 1.01 × 105 = 2.121 × 104 Pa (2) Suppose that the volume of the lungs remains unchanged, by the Pressure Law, the ratio between pressure and absolute temperature should be constant and hence the same for the partial pressure: P'/T' = P/T New partial pressure of oxygen in lungs: P' = 2.121 × 104 × (263/310) = 1.799 × 104 Pa
Chemistry(Gas Law)
發問:
Inspired air (breathed in) is about 21% oxygen, 0.93% argon, and the balance nitrogen. (The level of CO2 is negligible at <0.03%). If you lung capacity is 5.50 L at sea level on a standard day, what is the partial pressure of oxygen in your lung?If you climbed Mr. Rainier you would be at an elevation of... 顯示更多 Inspired air (breathed in) is about 21% oxygen, 0.93% argon, and the balance nitrogen. (The level of CO2 is negligible at <0.03%). If you lung capacity is 5.50 L at sea level on a standard day, what is the partial pressure of oxygen in your lung? If you climbed Mr. Rainier you would be at an elevation of 14000 ft above sea level, and the temperature would be -10.0 degree Celsius. What is the partial pressure of oxygen in your lungs at this elevation?
最佳解答:
Suppose that on a standard day and at the sea level, the atmospheric pressure is 1.01 × 105 Pa and the body temperature (in the lungs) is 310 K (37°C). Then by the ideal gas equation: PV = nRT n = PV/RT = (1.01 × 105 × 5.5 × 10-3)/(8.314 × 310) = 0.216 moles which is the total no. of moles of all gas in air. Then by Avogadro's law: Mole ratio is equal to volume ratio under the same temperature and pressure, we have: No. of moles of oxygen = 0.216 × 0.21 moles Further, by the Dalton's law: Partial pressure is proportional to the mole fraction, we have: Partial pressure of oxygen = [(0.216 × 0.21)/0.216] × 1.01 × 105 = 2.121 × 104 Pa (2) Suppose that the volume of the lungs remains unchanged, by the Pressure Law, the ratio between pressure and absolute temperature should be constant and hence the same for the partial pressure: P'/T' = P/T New partial pressure of oxygen in lungs: P' = 2.121 × 104 × (263/310) = 1.799 × 104 Pa
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