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States of matter. solids. liquids. gases. shape of container. independent. fixed shape. very small volumes. Kinetic model of gases. in constant motion. “elastic” collisions. Pressure. = force. force =. collisions of particles. area. 14.7 lbs. sea level. atmospheric pressure. - PowerPoint PPT Presentation
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States of matter
solids liquids gases
fixed shape shape of container independent
Kinetic model of gases very small volumesin constant motion
“elastic” collisions
Pressure = forcearea
atmospheric pressure
force =
14.7 lbsft2
sea level
collisions of particles
vacuum = absence of gas particles
atmospheric pressure barometer
760 mm Hg = 1 atm
760 mm Hg =760 torr
760 mm Hg =1.01325 x 105 Pa
could use water 10,300 mm H2O
34 feet
Torricelli
Boyle’s LawPressure Volume
PV = constant
Boyle’s LawPressure Volume
PV = constant
A sample of Cl2 has a volume of 946 mLat a pressure of 726 mm Hg.
If the volume is reduced to 154 mL, calculate the pressure.
P1V1 = P2V2
(726 mm Hg)(946 mL) = P2 (154 mL)
P2 = 4460 mm Hg
4460 mm Hg 760 mm Hg/atm = 5.97 atm
Charles’s LawVolume Temperature
VT
= constant
V = 0 T = -273.15oC
Kelvin
Charles’s LawVolume Temperature
V1
T1
= V2
T2
A container is filled with N2 in an ice bath (0o C)Its volume is 58.7 mL.
It is then put in a bath of boiling isopropanol.
Its volume goes to 76.4 mL.
What is the b.p. of isopropanol?
273.15 K
58.7 mL273.15 K
= 76.4 mL
T2
T2 = 355.5 K82.3o C
Avogadro’s LawVolume moles
Gay-Lussac 2 H2 + O2 2 H2O (g)
Law of combining volumes
200 mL H2 100 mL O2 200 mL H2O
Avogadro
equal volumes of gases = equal numbers of molecules
1 mol gas = 6.023 x 1023 particles = 22.4 L
at standard Temperature and Pressure0.0oC (273.2 K) 1 atm (760 torr)
Ideal Gas Law
PV = constantBoyle
Charles VT
= constant
Avogadro Vn
= constant
P V = n R T
R = ideal gas constant
= 0.08206 L atm
mol K
two types of problems
1. Predicting properties of a system PV = nRTno changes
2. Changing conditions P1V1
n1T1 n2T2
P2V2= R =
Calculate the pressure of 80.10 g. of CO2 gasin a 5000 mL flask at 70.0oC
1. Predicting properties of a system PV = nRT
P =
V = 5000 mL = 5.00 L
n = 80.10 g ÷ 44.01 g/mol = 1.82 mol
T = 70.2 + 273.15 = 343.2 K
(P)mol K
(5.00 L) = (1.82 mol)(0.08206 L atm)( 343.2 K)
(P) = 10.25 atm
A small bubble (2.1 mL) rises from the bottom of a lake,where the temperature is 8.00Cand the pressure is 6.4 atm, to the surface
where the temperature is 25.0oC.
What is the volume of the bubble at the surface?
2. Changing conditions = R =P1V1
n1T1 n2T2
P2V2
P1= 6.4 atm
V1 = 2.1 x 10-3 L
T1 =
initial
281.15 K
final
P2= 1.0 atm
V2 =
T2 = 298.15 K
14 mL