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## Three States of Matter

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### First mercury barometer made by Evangilista Torricelli in 1643. The weight of the column of mercury is equal to the force exerted by the atmosphere ... – PowerPoint PPT presentation

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Title: Three States of Matter

1
Three States of Matter
2
Physical Properties of Gases (Page 161)
Because of the great distance between particles,
gas behavior is fundamentally different from the
other phases
Gases assume the volume and shape of their
container
Gases have lower densities and are more
compressible than solids and liquids
Gases will mix uniformly when placed in the same
container
3
The Gaseous Elements
(at room temperature)
Common Gaseous Compounds
NH3 ammonia
HCl hydrogen chloride
CO2 carbon dioxide
SO2 sulfur dioxide
Be sure you are familiar with all of the gaseous
substances listed in table 5.1 on page 161
4
Pressure
Gases exert pressure on objects in their
surroundings.
Pressure is caused by collisions between gas
particles and the objects with which they are in
contact.
Pressure is the force exerted on a unit of area.
Pressure Force
area
5
Atmospheric pressure is caused by the weight of
air in the earths atmosphere
Atmospheric pressure pushes equally in all
directions
It is measured with a barometer.
6
Making a Barometer
7
Pressure Force
area
First mercury barometer made by Evangilista
Torricelli in 1643
The weight of the column of mercury is equal to
the force exerted by the atmosphere
Normal atmospheric pressure
760 mm of Hg
8
Units for Normal Atmospheric Pressure
Patm Normal Atm Pressure 760 mm Hg.
Patm 760 torr.
1.00 mm Hg 1.00 torr.
Patm 1.00 atm 760 torr.
Patm 14.7 lbs/in2
9
SI Units for Normal Atmospheric Pressure
Pressure Force
area
1 pascal 1 newton of force
1 meter 2
Patm 101.3 kilopascals (kPa)
Patm 760 mm Hg 101.3 kPa
10
The Manometer
A manometer measures gas pressure as a difference
in Hg column heights.
Closed-end manometer difference in column
heights gives absolute pressure.
Open-end manometer difference in column heights
gives difference between gas sample and
atmospheric pressures
11
Open-end Manometer
Patm gt Pgas
Pgas h
Patm
Patm - h
Pgas
748 mm Hg - 78 mm Hg
Pgas
670 mm Hg
Pgas
12
Open-end Manometer
Pgas gt Patm
Patm h
Pgas
748 mm Hg 57 mm Hg
Pgas
805 mm Hg
Pgas
13
Closed-end Manometer
Pgas gt 0
h 0 torr
Pgas
85 mm Hg
Pgas
14
The Four Gas Parameters
• Four parameters are needed to define the
physical condition or state of any gas
• Temperature (T)
• Pressure (P)
• Volume (V)
• Amount of gas (moles n)
• Equations relating these variables are known as
the gas laws.

15
The Gas Laws
Boyles Law (P-V Relationship)
Charles Law (T-V Relationship)
Avogadros Law (n-V Relationship)
Gay-Lussacs Law (P-T Relationship)
The two parameters not being studied are held
constant
16
Boyles Law
Temperature and amount remain constant
17
Boyles Law
P a 1/V
P K /V
PV K
Temperature and amount remain constant
P1V1 P2V2
18
BOYLES LAW (P-V RELATIONSHIP)
The volume of a fixed amount of gas maintained at
constant temperature is inversely proportional to
pressure.
(T and n constant)
19
Charless Law
V a T
V K T
V/T K
Pressure and amount remain constant
20
• Charless Law and Absolute Zero

K oC 273
oC K - 273
Absolute zero is the temperature at which there
is no energy.
-273.15oC
21
Gay-Lussacs Law
P a T
P K T
P/T K
Volume and amount remain constant
22
V a n
V K n
V/n K
Pressure and temperature remain constant
23
Ideal Gas Law
Boyles Law
Charless Law
Expressed Mathematically
24
Finding the Value of R
Standard temperature 273.15 K (0C) Standard
Pressure 1 atm
At STP 1 mole of any gas occupies 22.414 L
solving
gives
25
Some Values of the Gas Constant R
R 0.08206 Latm
mol K
R 62.36 Ltorr
mol K
R 8.31 LkPa
mol K
26
Single State Problem
A 12.25 L cylinder contains 75.5 g of neon at
24.5 oC. Determine the pressure in the cylinder.
What type of problem is this?
Only one set of conditions
27
A 12.25 L cylinder contains 75.5 g of neon at
24.5 oC. Determine the pressure in the cylinder.
P nRT V
PV nRT
(3.74 mol)(62.4Ltorr)(297.5K) (12.25 L)
molK
?
P V n R T
12.25 L
mol
5667.7 torr
75.5 g mol
3.74
20.18 g
5670 torr
62.4 Ltorr molK
How many atmospheres is this?
24.5 273 297.5 K
28
Double State Problem
A balloon contains helium gas with a volume of
2.60 L at 25 oC and 768 mmHg. If the balloon
ascends to an altitude where the helium pressure
is 590 mmHg and the temperature is 15 oC, what is
the volume of the balloon?
What type of problem is this?
There are 2 sets of conditions.
29
A balloon contains helium gas with a volume of
2.60 L at 25 oC and 768 mmHg. If the balloon
ascends to an altitude where the helium pressure
is 590 mm Hg and the temperature is 15 oC, what
is the volume of the balloon?
P1 V1 T1 n1
P2 V2 T2 n2
768 torr
590 torr
2.60 L
?
25 273 298 K
15 273 288 K
(768 torr)(2.60 L)(288 K) (590
torr)(298 K)
3.27 L
30
A 2.50 gram sample of a solid was vaporized in a
505 mL vessel. If the vapor pressure of the
solid was 755 mmHg at 155 oC, what is the
molecular weight of the solid?
molecular weight molar mass g/mol g ? mol
..so if we can find grams and moles and divide....
...we already have grams!! Were halfway there!
P V n R T
755 torr
n PV RT
0.505 L
755 torr 0.505 L molK___________
62.4 Ltorr
428 K
62.4 Ltorr molK
0.01428 mol
155 273 428 K
NOW 2.50 g g
0.01428 mol mol
175.1