Transcript The Behavior of Gases Unit 11

The Behavior of Gases
Unit 9
GENERAL CHEMISTRY
SPRING 2010
Mr. Hoffman
Mrs. Paustian
Air Pressure
 Pressure equation
Newtons (N)
Square meters (m2)
Unit for Air Pressure
f
P
a
 Pascals
1 kPa = 1000
 1 N/m2 = 1 Pascal (Pa)
Pa!
 Standard pressure is 101.325 kPa on Earth
Blaise Pascal
(French philosopher and
mathematician)
Pascal’s Triangle
Air Pressure
 Without changing the
mass (force), how can
you increase pressure?
Decrease the area
f
P
a
 Without changing the
mass (force), how can
you decrease pressure?
Increase area
Practice Problem
f
P
a
 The mass of a brick is 2.0 kg (F=19.6N), the sides are
0.05m and 0.03m. What is the pressure exerted?
19.6 N
(0.05 m x 0.03 m)
= 13,067 Pa
Pressure Units
Units of Gas Pressure
 Ways to represent
pressure
Torr is named
after
Evangelista
Torricelli
Unit
Standard Pressure
Atmosphere
1 atm (exactly)
Inches of Hg
29.9 in Hg
cm of Hg
76 cm Hg
mm of Hg
760 mm Hg
Torr
760 torr
Pounds per sq. in.
14.7 psi
Kilopascal
101 kPa
The Barometer
 Etymology of “barometer”
 In Greek, “baros”= weight
 Meter= measure
 Literally means “measure the weight of air” or air pressure.
Bariatric
surgeory is
weight loss
surgeory
$5 Footlong
The Barometer
 How a barometer works



Air presses down on an
open tray of Hg
This downward pressure
pushes the Hg up into a
tube
Higher air pressure causes
the mercury to go higher up
in the tube (measured as
height, mm Hg)
Compressibility
 A gas will expand to fill its
container
 Compressibility


A measure of how much the
volume of matter decreases
under pressure.
Gases easily compress because
of the space between the
particles
http://www.garagelibrary.com/images/airbag.jpg
Kinetic Molecular Theory
 Postulates the theory is based on…
1.
2.
3.
4.
Gases consist of tiny particles (of negligible mass) with
great distances separating them
Gases are in constant random motion
Molecular collisions are elastic
Average kinetic energy is dependent only on
temperature
The Kelvin Scale
 As T increases, so does kinetic energy
 Theoretically, kinetic energy can be zero, but
it hasn’t been achieved and probably won’t
ever be achieved
 Absolute zero- The temperature at which a
substance would have zero kinetic energy
 The Kelvin Scale- a temperature scale directly
related to kinetic energy
 Zero
on the Kelvin scale corresponds to zero kinetic
energy
The Kelvin Scale
 Units are Kelvins (K), with no degree (o) sign
Kelvin relates
temperature
to kinetic
energy!
Temperature Conversions
 Easy to convert between
Celsius and Kelvin

How do you think?
 oC
 K? Add 273
 K  oC? Substract 273

25oC  K?


(25+273)= 298 K
310 K  oC?

(310-273) = 37oC
 Fahrenheit Celsius?
 (oF – 32oF) x 5/9 = oC
 (oC x 9/5) + 32oF = oF
Factors Affecting Gas Pressure
 Amount of gas
 Volume
 Temperature
http://www.bmumford.com/photo/highspeed/Ted1.jpg
Boyle’s Law
 “Boyles inverse”
 States that the volume of a gas varies inversely
(opposite) with pressure if temperature and
amount are held constant
 Written:
 P1V1
= P2V2
Robert Boyle
Boyle’s Law in motion
http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/gaslab/Images/chprmt.gif
Boyle’s Practice 1
 A tank contains a volume of 3 L and a
pressure of 4 atmospheres. What volume
would the gas from this tank fill up at a
pressure of 1 atmosphere?
 P1
= 4atm
 V1 = 3L
 P2 = 1 atm
 V2 = ?
Substitute into Boyle’s Law equation and solve for V2
P1V1 = P2V2
(4 atm)(3L) = (1atm)(V2)
12 atm*L = V2
1 atm
V2= 12 L
Boyle’s Practice 2
 Find the volume of a cylinder needed if you
want to put 50 atmospheres of pressure with a
volume of 3 L into a cylinder that can hold a
pressure of no greater than 20 atmospheres.
 P1=
50 atm
 V 1= 3 L
 P2= 20 atm
 V2= ?
Substitute into Boyle’s Law equation and solve for V2
P1V1 = P2V2
(50 atm)(3 L) = (20 atm)(V2)
7.5 L = V2
Charles’ Law
 “Charles direct”
 The volume of a gas varies directly with
temperature if the pressure and amount
remain constant
 Mathematically:
Charles’ Law in Motion
Charles’ Law Problem 1
 Always convert temperature to Kelvin (K)
 A tank contains a volume of 3 L and a temperature of
100oC. What volume would the gas from this tank fill
up at a temperature of 200oC?




V1= 3 L
T1= 100oC = 373K
V2= ?
T2= 200oC= 473K
Substitute into equation and
solve for V2
3L
V2
=
373K
473K
(3L)(473K) = (V2)(373K)
4L
=
V2
Charles’ Law Problem 2
 A 275 L helium balloon is heated from 20oC to 40oC.
Calculate the final volume assuming pressure
remains constant.




V1= 275 L
T1= 20oC (+273K)= 293K
V2= ?
T2= 40+273= 313K
275L
293K
= V2
313K
(293)(v2) = (275L)(313K)
V2= 294 L
Temperature-Pressure Relationships
 Gay-Lussac’s Law
 The pressure of a gas varies directly w/ temperature if volume
and amount remain constant.
 Mathematically:
Tire inflation….
Summer vs.
winter…hmmm how
do these two differ?
Joseph Louis Gay-Lussac
1778-1850
Gay-Lussac Problem 1
 Temperature in Kelvin (K)
 A tank contains a pressure of 3 atm and a
temperature of 100oC. What pressure would the gas
from this tank be at a temperature of 200oC?




P1= 3 atm
T1= 100oC (+273K)= 373K
P2= ?
T2= 200oC= 473K
3 atm = P2
373 K
473 K
P2= 4 atm
Gay-Lussac Problem 2
 Find the pressure needed if you wanted to put gas at
50oC and 75 atm into a vessel that is at 65oC.




P1= 75 atm
T1= 50oC= 323 K
P2= ?
T2= 65oC= 338 K
75 atm = P2
323 K 338 K
P2 = 78 atm
Combined Gas Law
 Puts together several scientists’ work on how gases
behave when conditions are changes.
 You can change three things about a gas




Amount of gas- this stays the same for now
Temperature
Volume
STP??
Temperature= 273K
Pressure changes in response to these
Pressure= 1 atm
CBL Problem 1
 A 50 mL sample of hydrogen gas is collected at 772
mm Hg and 21oC. Calculate the volume of hydrogen
at STP.






P1= 772 mm Hg  atm
V1= 50 mL  L
T1= 21oC= 294 K
P2= 1 atm
V2= X
T2= 273K
Since your final
conditions are at STP,
you need to convert initial
conditions to atm and L
CBL Problem 1 solution
P1= 772 mm Hg x 1 atm/760 mm Hg= 1.016 atm
V1= 50 mL / 1000 mL= 0.05 L
(1.016 atm)(0.05L) = (1atm)(V2)
294 K
273 K
Cross multiply, then solve for V2
(1.016 atm)(0.05L)(273 K) = (1atm)(294K)(X)
0.05 L = V2
Ideal Gases
 Ideal gases are…
 Gases that behave under all conditions as
predicted by the kinetic molecular theory
 Gases are not ideal…
 When they don’t behave as predicted by kinetic
molecular theory
 Kinetic energy is…
 Energy associated with motion
The Ideal Gas Law
 “piv-nert”
 PV=nRT
 P= Pressure (atm)
 V= Volume (L)
 n= # moles (mol)
 R= Ideal gas constant= 0.0821 (L*atm)/(mol*K)
 T= Temperature (ALWAYS Kelvin)
IGL Problem 1
 If a container has a volume of 3 L and is at a
temperature of 60oC and a pressure of 6 atm, how
many moles of gas are in the container?





V= 3L
T= 60oC= 333K
P= 6 atm
R= 0.0821 L*atm/mol*K
n= ?
IGL Problem 1 Solution
 Substitute in PV=nRT and solve for n
(6 atm)(3 L) = (n)(0.0821)(333 K)
n= 0.7 mol
The unit for R is
L*atm
mol*K
Be sure units match this,
and units will cancel
IGL Problem 2
 If a container has 50 moles of O2 and is at a
temperature of 40oC and a pressure of 3 atm, how
many liters are in this container?





n= 50 mol O2
T= 40oC = 313 K
P= 3 atm
R= 0.0821 L*atm/mol*K
V= ?
IGL Problem 2 Solution
 Substitute in PV=nRT and solve for V
(3atm)(V) = (50 mol)(0.0821)(313 K)
V = 428 L