I: Partial pressure of Nitrogen
Ambient pressure is the total pressure = 745mmHg.
Vapour pressure of water = 16.48mmHg
According to Dalton's Law of partial pressure;
P(Total) = P1 + P2 + .......
Hence;
P(Total) = PN2 + PH2O
745mmHg = PN2 + 16.48mmHg
PN2 = (745 - 16.48) mmHg
= 728.52mmHg
II: Moles of water in wet gas
Given that;
PV = nRT
and;
P = 16.48mmHg
V = 500mL = 500x10-3L
R = 62.36367LmmHgK-1mol-1
T = 19oC + 273 = 292K
n=?
Therefore;
n"= \\dfrac{PV}{RT} = \\dfrac{16.48mmHgx 500x10^-3L}{62.36367LmmHgmol^-1K^-1x292K}" = 4.525x10-4 mol
III: Moles of dry gas collected
Since; PV=nRT
and;
P= 728.52mmHg
V= 500x10-3L
T = 292K
n "=\\dfrac{PV}{RT} = \\dfrac{728.52mmHgx500x10^-3L}{62.36367LmmHgmol^-1K^-1x292K} =0.02mol"
IV: Partial pressure of Neon
Neon added = 0.128g
Moles of neone "\\dfrac{mass}{Molar mass} = \\dfrac{0.128g}{20.18g\/mol}" = 6.34 x 10-3 mol
Since PV=nRT
where n=6.34 x 10-3 mol
R = 62.36367LmmHgK-1mol-1
T = 19oC + 273 = 292K
V= 0.5L
P "=" "\\dfrac{nRT}{V} = \\dfrac{6.34x10^-3molx62.36367LmmHgK^-1K^-1x292K}{0.5L}" = 230.9mmHg
V: Total pressure after Neon is added
P(total) = PH2O + PN2 + PNe
= (16.48 + 728.52 + 230.9)mmHg
=975.9mmHg
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