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Question-Partial pressure

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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? 更新: Thanks you

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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) =

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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 0.523m30D7DAC4E7B8CAAC5

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