DATA Temperature of boiling water Temperature of cool water Barometric pressure
ID: 1026450 • Letter: D
Question
DATA Temperature of boiling water Temperature of cool water Barometric pressure Vapor pressure of water Partial pressure of cool gas 140 Volume of water in flask Trial 1 Trial 2 slip # 8,36 2.56 inches Wa 2q.S0 iackes H 14o,3 torr un.b w 283.4-mL 63 tor 61.51. Volume of flask with stopper Volume of the cool gas 1.4 mu Data partner for Trial 1: (remember, you may not work together on the report) Record data in the units actually measured (e.g., °C). It would be acceptable to also include the value converted into another unit (e.g., 18.0°C- 291.2 K). Each experimenter is responsible to do the unit conversions correctly even when data are sharedExplanation / Answer
1) trial 1)
P1=pressure of cool gas=740.63 torr
T1=temperature of cool gas=9.1+273=282.1 K
V1=volume of cool gas=288ml=0.288L
P2=pressure of hot gas=Patm-P(water vapor)=29.5torr-8.67torr=20.83 torr
V2=volume of hot gas
T2=99.4+273=372.4K
Using eqn, P1V1/T1=P2V2/T2
or, (740.63torr)(288ml)/(282.1K)=(20.83torr)*V2/(372.4K)
V2=13517.941 ml=13.518L
trial 2)
P1=pressure of cool gas=740.63 torr
T1=temperature of cool gas=18.8+273=291.8 K
V1=volume of cool gas=241.4ml
P2=pressure of hot gas=Patm-P(water vapor)=29.5torr-8.67torr=20.83 torr
V2=volume of hot gas
T2=99.4+273=372.4K
Using eqn, P1V1/T1=P2V2/T2
or, (740.63torr)(241.4ml)/(291.8K)=(20.83torr)*V2/(372.4K)
V2=(740.63torr)(241.4ml)(372.4K)/(291.8K)(20.83torr)
V2=10954.024=10.954L
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