states of matter

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