A mixture containing 0.516 mol He(g), 0.310 mol Ne(g), and 0.150 mol Ar(g) is confined in a 8.00-L vessel at 22°C.
(a) Calculate the partial pressure of each of the gases in the mixture.
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2018-01-24 10:24 am
回答 (1)
2018-01-24 11:09 am
For the gas mixture :
Total number of moles, n = (0.516 + 0.310 + 0.150) mol = 0.976 mol
Volume, V = 8.00 L
Temperature, T = (273 + 22) K = 295 K
Gas constant, R = 0.08206 L atm / (mol K)
Gas law: PV = nRT
Total pressure, P = nRT/V = 0.976 × 0.08206 × 295 / 8.00 atm = 2.95 atm
Partial pressure = (Total pressure) × (Mole fraction)
Partial pressure of He, P(He) = (2.95 atm) × (0.516/0.976) = 1.56 atm
Partial pressure of Ne, P(Ne) = (2.95 atm) × (0.310/0.976) = 0.937 atm
Partial pressure of Ar, P(Ar) = (2.95 atm) × (0.150/0.976) = 0.453 atm
Total number of moles, n = (0.516 + 0.310 + 0.150) mol = 0.976 mol
Volume, V = 8.00 L
Temperature, T = (273 + 22) K = 295 K
Gas constant, R = 0.08206 L atm / (mol K)
Gas law: PV = nRT
Total pressure, P = nRT/V = 0.976 × 0.08206 × 295 / 8.00 atm = 2.95 atm
Partial pressure = (Total pressure) × (Mole fraction)
Partial pressure of He, P(He) = (2.95 atm) × (0.516/0.976) = 1.56 atm
Partial pressure of Ne, P(Ne) = (2.95 atm) × (0.310/0.976) = 0.937 atm
Partial pressure of Ar, P(Ar) = (2.95 atm) × (0.150/0.976) = 0.453 atm
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