A gas mixture with a total pressure of 760 mmHg contains each of the following gases at the indicated partial pressures: 119 mmHg CO2, 211 mmHg Ar, and 186 mmHg O2. The mixture also contains helium gas.
1) What is the partial pressure of the helium gas? Answer: 244 mmHG
2) What mass of helium gas is present in a 13.3-L sample of this mixture at 264 K ? (need this answer)
What mass of helium gas is present in a 13.3-L sample of this mixture at 264 K ?
2020-04-27 10:38 pm
回答 (1)
2020-04-27 11:36 pm
1)
P(total) = P(CO₂) + P(Ar) + P(O₂) + P(He)
760 mmHg = (119 mmHg) + (211 mmHg) + (186 mmHg) + P(He)
Partial pressure of He, P(He) = (760 - 119 - 211 - 186) mmHg = 244 mmHg
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2)
For the He gas:
Pressure, P = 244/760 atm
Volume, V = 13.3 L
Mass, m = ? g
Molar mass of He, M = 4.003 g/mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = 264 K
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, m = PVM/(RT)
Mass of He, m = (244/760) × 13.3 × 4.003 / (0.08206 × 264) g = 0.789 g
P(total) = P(CO₂) + P(Ar) + P(O₂) + P(He)
760 mmHg = (119 mmHg) + (211 mmHg) + (186 mmHg) + P(He)
Partial pressure of He, P(He) = (760 - 119 - 211 - 186) mmHg = 244 mmHg
====
2)
For the He gas:
Pressure, P = 244/760 atm
Volume, V = 13.3 L
Mass, m = ? g
Molar mass of He, M = 4.003 g/mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = 264 K
PV = nRT and n = m/M
Then, PV = (m/M)RT
Hence, m = PVM/(RT)
Mass of He, m = (244/760) × 13.3 × 4.003 / (0.08206 × 264) g = 0.789 g
收錄日期: 2021-04-18 18:32:59
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