a) How many molecules of water are in 1.00 cm^3 of saturated air at 20. degree celcius?
b) What volume of saturated air at 20. degree celcius contains 0.500 mol of water?
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Question-Partial pressure
2009-05-31 9:03 am
The partial pressure of water vapour in saturated air at 20. degree celcius is 0.0230 atm.
a) How many molecules of water are in 1.00 cm^3 of saturated air at 20. degree celcius?
b) What volume of saturated air at 20. degree celcius contains 0.500 mol of water?
a) How many molecules of water are in 1.00 cm^3 of saturated air at 20. degree celcius?
b) What volume of saturated air at 20. degree celcius contains 0.500 mol of water?
回答 (2)
2009-06-01 5:25 am
✔ 最佳答案
a)Pressure P = 0.023 atm = 0.023 x 101000 Pa
Temperature T = 20 + 273 = 293 K
Volume V = 1 cm3 = 1 x 10-6 m3
Gas constant R = 8.31 J mol-1 K-1
Avogadro constant = 6.02 x 1023 mol-1
No. of moles n = ? mol
PV = nRT
n = PV/RT
n = (0.023 x 101000)(1 x 10-6)/(8.31)(293)
No. of water molecules
= [(0.023 x 101000)(1 x 10-6)/(8.31)(293)] x (6.02 x 1023)
= 5.74 x 1017 molecules
b)
Pressure P = 0.023 atm = 0.023 x 101000 Pa
Temperature T = 20 + 273 = 293 K
No. of moles n = 0.5 mol
Gas constant R = 8.31 J mol-1 K-1
Volume V = ? m3
PV = nRT
V = nRT/P
V = (0.5)(8.31)(293)/(0.023 x 101000)
V = 0.524 m3
Volume of saturated air = 0.524 m3 (or 524 dm3)
=
2009-05-31 12:25 pm
use PV=nRT
assume is an ideal gas
partial pressure of H2O = 0.023atm
R= 82.06cm^3atm/gmolK or 8.314kj/kmolK
T= 273.15+20=293.15K
a)
0.023*1=n*82.06*293.15
n= 9.56*10^-7gmol
b) if 0.5mol =0.5gmol
nRT/P=V
0.5*82.06*293.15/0.023=V =5.23*10^5cm3 or 0.523m3
assume is an ideal gas
partial pressure of H2O = 0.023atm
R= 82.06cm^3atm/gmolK or 8.314kj/kmolK
T= 273.15+20=293.15K
a)
0.023*1=n*82.06*293.15
n= 9.56*10^-7gmol
b) if 0.5mol =0.5gmol
nRT/P=V
0.5*82.06*293.15/0.023=V =5.23*10^5cm3 or 0.523m3
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