Gas Laws Day 3. Gas Law Foldable Fold the left and right to the middle.
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Transcript of Gas Laws Day 3. Gas Law Foldable Fold the left and right to the middle.
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Gas Laws
Day 3
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Gas Law Foldable
Fold the left and right to the middle.
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Dalton’s Law of Partial Pressure
The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
PTotal = P1 + P2 + P3….
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Example
A balloon is filled with air (O2,
CO2, & N2) at a pressure of
1.3 atm.
If PO2 = 0.4 atm and PCO2 =
0.3 atm, what is the partial pressure of the nitrogen gas?
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PTotal = P1 + P2 + P3….
Ptotal = PO2 + PCO2 + PN2
1.3 atm = 0.4 atm + 0.3 atm + PN2
PN2 = 0.6 atm
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Boyle’s Law: Pressure vs. Volume
At a constant temperature, the volume of a fixed mass of gas varies inversely with the pressure
P1V1 = P2V2
*use if temperature is constant
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Boyle's Law
0
500
1000
1500
1 2 3 4 5 6 7
Pressure (atm)
Vo
lum
e (
mL
)
InverseIndirect
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Pressure and Volume with constant Temperature
Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.
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Uses of Boyle’s LawTesting materials stability and ability to
maintain their shape under force.Compressing gases for use in cooking
cylinders, SCUBA tanks, and shaving cream.Used to describe density relationships
between gases.
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ExampleIf a gas expands from a volume of 5 L
to 25 L at an initial pressure of 3.5 atm, what will be its new pressure?
P1 = 3.5 atm
V1= 5 L
P2 = ?
V2 = 25 L
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Charles’ Law: Volume vs Temperature
At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature
V1 = V2
T1 T2
*use if pressure is constant
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Charles' Law ( Vol v. Temp)
0
200
400
600
800
1000
1200
0 200 400 600
Volume (mL)
Te
mp
era
ture
(K
) Direct
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Charles’ Law Graph
Temp decreases
Vol decreases.
Temperature (K)
Volume (mL)
546 1092
373 746
283 566
274 548
273 546
272 544
200 400
50 100
0 0
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Temperature and Volume with constant Pressure
Changing the temperature but requiring the pressure to stay the same causes the volume to increase.
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Uses of Charles’ Law
Describing the properties of gases, liquids, and solids at extremely low temperatures.
Hot air ballooningUsed to describe density
relationships of gases.
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Example
If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume?
V1 = 22.4 L
T1 = 23 + 273 = 296 K
V2 = ?
T2 = 50 + 273 = 323 K
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Absolute Zero
Temperature at which all molecular motion stops.
It is defined by 0 K or -273C. Scientists used Charles Law to
extrapolate the temperature of absolute zero.
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Avogadro’s Law: Volume vs. Moles
At a constant temperature & pressure, the volume of a gas varies directly with the moles
V1 = V2
n1 n2
*use if temperature and pressure are constant
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Avogadro’s Law
As the number of moles increases, the volume expands to make room for the additional gas
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Example
If 4.65 L of CO2 increases from 0.8 moles to 3.75 moles, what is the new volume of the gas?
V1 = 4.65 L
n1 = 0.8 moles
V2 = ? L
n2 = 3.75 moles
V1 = V2
n1 n2
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V2 = V1n2
n1
V2 = (4.65 L)(3.75 mol) = 21.79 L
(0.8 mol) (20 L)
V1 = V2
n1 n2
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The Combined Gas Law
P1 V1 = P2 V2
n1T1 n2T2
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The Combined Gas LawExpresses a relationship between
pressure, volume, and temperature (and moles) of a fixed amount of gas.
It takes all three gas laws: (Boyle’s,
Charles’s, & Avogadro’s) and combines them form one usable equation.
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Example 1A gas is cooled from 45˚C to 20˚C. The
pressure changes from 103 kPa to 101.3 kPa as the volumes settles to 16.0 L. What was the initial volume?
P1 = 103 kPa
V1 = ?
T1 = 45oC + 273 = 318 KP2 = 101.3 kPa
V2 = 16.0 L
T2 = 20oC + 273 = 293 K
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(103 kPa)(V1) = (101.3 kPa)(16.0 L)
(n1) (318 K) (n2) (293 K)
V1 = 17.1 L
P1 V1 = P2 V2
n1T1 n2T2
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Example 2 (on back of foldable)A 1.5 mole sample of methane was originally
0.5 L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C?
P1 = 1.1 atm
V1 = 0.5 L
T1 = 25 + 273= 298 K
n1 = 1.5 moles
P2 = 2.0 atm
V2 = 0.25 L
T2 = ?
n2 = 1.5+2.5 = 4 moles
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T2 = 101.59 K – 273
= -171.41 ˚C
P1 V1 = P2 V2
n1T1 n2T2*** easier if you solve for T2 first THEN plug in the #’s ***
P1 V1n2T2 = P2 V2n1T1
P1 V1n2 P1 V1n2
= (2.0 atm)(0.25 L)(1.5 mol)(298K)
(1.1 atm)(0.5 L)(4.0 mol)