A steel cylinder holds 1.52 g of ethanol, C2H5OH. What is the pressure of the ethanol vapor if the cylinder has a volume of 237 cm3 and the temperature is 267°C? (Assume all of the ethanol is in the vapor phase at this temperature.)
P = atm
chem help?!?
2016-06-22 8:42 pm
回答 (2)
2016-06-22 9:00 pm
C₂H₅OH gas :
Pressure, P = ? atm
Volume, V = 237 cm³ = 0.237 L
Mass, m = 1.52 g
Molar mass, M = (14.01×2 + 1.008×6 + 16.00) g/mol = 50.07 g/mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 267) K = 540 K
PV = nRT
PV = (m/M)RT
P = mRT/(MV)
Pressure, P = 1.52 × 0.0821 × 540 / (50.07 × 0.237) = 5.68 atm
Pressure, P = ? atm
Volume, V = 237 cm³ = 0.237 L
Mass, m = 1.52 g
Molar mass, M = (14.01×2 + 1.008×6 + 16.00) g/mol = 50.07 g/mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 267) K = 540 K
PV = nRT
PV = (m/M)RT
P = mRT/(MV)
Pressure, P = 1.52 × 0.0821 × 540 / (50.07 × 0.237) = 5.68 atm
2016-06-22 8:42 pm
Hmm very high
收錄日期: 2021-04-18 15:10:27
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