Unit for Air Pressure
Pascals 1 N/m2 = 1 Pascal (Pa) Standard pressure is 101.325 kPa on Earth
f
P a
Blaise Pascal(French philosopher and mathematician)
Pascal’s Triangle
1 kPa = 1000 Pa!
Air Pressure
Without changing the mass (force), how can you increase pressure?
Without changing the mass (force), how can you decrease pressure?
f
P a
Decrease the area Increase area
Practice Problem
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?
f
P a
19.6 N = 13,067 Pa
(0.05 m x 0.03 m)
Pressure Units
Ways to represent pressure
Units of Gas PressureUnit 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
Torr is named after
Evangelista Torricelli
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. Gases consist of tiny particles (of negligible
mass) with great distances separating them2. Gases are in constant random motion3. Molecular collisions are elastic
4. Average kinetic energy is dependent only on temperature
The Kelvin Scale
As T increases, so does kinetic energyTheoretically, 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 gasVolumeTemperature
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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 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 atmV2= 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 V1= 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 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 = V2
293K 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:
Joseph Louis Gay-Lussac 1778-1850
Tire inflation…. Summer vs. winter…hmmm how do these
two differ?
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 Pressure changes in response to these
STP??Temperature= 273K
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
(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
P1= 772 mm Hg x 1 atm/760 mm Hg= 1.016 atm
V1= 50 mL / 1000 mL= 0.05 L
Ideal Gases
Ideal gases are…Gases that behave under all conditions as
predicted by the kinetic molecular theoryGases 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= ?