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Whitney's lung capacity was measured as 3.2 l at a body temperature of 37 ∘c and a pressure of 747 mmhg . how many moles of oxygen is in her lungs if air contains 21% oxygen?

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Kalahira
Whitney has 0.026 moles of O2 in her lungs. The ideal gas law is PV = nRT where P = Pressure V = volume n = Number of moles R = Ideal Gas Constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature Since a mmHg is 1/760 atm, the pressure is 747/760 atm = 0.982894737 atm Converting 37C to K, 37 + 273.15 = 310.15 K Now let's take the ideal gas equation, solve for n, the substitute the known values and solve. PV = nRT PV/RT = n (0.982894737 atm)(3.2 L)/((0.082057338 L*atm/(K*mol))310.15 K) = n (3.145263158 atm*L)/(25.45008338 L*atm/mol) = n 0.123585574 mol = n So there's 0.123585574 moles of gas particles in Whitney's lungs. Let's get the moles of oxygen 0.123585574 mol * 0.21 = 0.02595297 mol Rounding to 2 significant figures gives 0.026 moles.