Question-Partial pressure

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)
=

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