In: Chemistry
50 L of a gas were collected over water when the barometer read 684.0 mmHg and the temperature was 14 oC . What volume would the dry gas occupy at standard conditions?
HINT: Consider Dalton’s law of partial pressures.
PT = 648mmHg
PH2O = 12mmHg ( water vapor pressure at 140C)
Pgas = PT-PH2O
= 684-12 = 672mmHg
Given conditions STP conditions
V1 = 50L V2 =
T1 = 14+273 = 287K T2 = 273K
P1 = 672mmHg P2 = 760mmHg
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2
= 672*50*273/287*760 = 42L
The volume of gas = 42L