The ideal gas law

Report
Unit 1 Gases
The Nature of Gases
Objectives:
1. Describe the assumption of the kinetic theory as it
applies to gases.
2. Interpret gas pressure in terms of kinetic theory
3. Define the relationship between Kelvin temperature
and average kinetic energy.
4. Explain why gases are easier to compress than solids
or liquids.
5. Describe the 3 factors that affect gas pressure.
 The Kinetic Theory
 Properties of Gases
 Expand to fill their container
 take the shape of their container
 low density
 Compressible
• Compressibility measures how much the
volume of matter decreases under pressure.
 mixtures of gases are
always homogeneous
 Fluids (flow)
 Gas pressure
 Results from collisions of gas particles with an object.
 In empty space where there are no particles, there is no
pressure and is called a vacuum.
 Atmospheric pressure (air pressure): due to atoms and
molecules in air.
 Barometer: used to
measure atmospheric
pressure.
 Units for measuring pressure:



Pascal (Pa)
Standard atmosphere (atm)
Millimeters of mercury (mmHg)
1 atm = 760 mmHg = 101.3 kPa
 1kpa = 1000 pa
 Standard pressure: 1 atm

 Factors affecting gas pressure
 Amount of gas
 Volume
 Temperature

Standard temperature : 0C (273K)
 Converting between units of pressure
A pressure gauge records a pressure of 450 kPa.
What is the measurement expressed in atmospheres
and millimeters of mercury?
For converting to atm:
450 kpa x 1 atm = 4.4 atm
1013.kPa
For converting to mmHg:
1.
450kPa x 760 mmHg = 3.4 x 103 mmHg
101.3 kPa
2. What pressure in kilopascals and in atmospheres,
does a gas exert at 385 mmHg?
3. The pressure on the top of Mount Everest is 33.7 kPa.
Is that pressure greater or less than 0.25atm?
2. What pressure in kilopascals and in atmospheres,
does a gas exert at 385 mmHg?
51.3 kPa, 0.507 atm
3. The pressure on the top of Mount Everest is 33.7 kPa.
Is that pressure greater or less than 0.25atm?
33.7 kPa is greater than 0.25 atm
 Kinetic energy and temperature
 The Kelvin temperature of a substance is directly
proportional to the average kinetic energy of the
particles of the substance.
 Directly proportional means that temperatures
increases as the average kinetic energy increases or
decreases as the average kinetic energy decreases.
 K = C + 273
The Atmosphere
 an “ocean” of gases
mixed together
Composition
nitrogen (N2)…………..
~78%
oxygen (O2)……………
~21%
argon (Ar)……………...
~1%
Trace amounts of:
carbon dioxide (CO2)…
water vapor (H2O)…….
~0.04%
~0.1%
He, Ne, Rn, SO2,
CH4, NxOx, etc.
Depletion of the Ozone Layer
O3 depletion is caused by chlorofluorocarbons (CFCs).
Ozone (O3) in upper atmosphere
blocks ultraviolet (UV) light from Sun.
UV causes skin cancer and cataracts.
Uses for CFCs:
refrigerants
CFCs
aerosol propellants
-- banned in U.S. in 1996
O3 is replenished with each strike of lightning.
 Reaction_to_Air_Pressure_Below_Sea_Level.asf
 Classwork:
Read pages 103-105
Do problems 1,4,5,6
Gas Laws
 Objectives
Describe the relationships among the temperature,
pressure, and volume of a gas
2. Use the gas laws to solve problems
1.
Boyle’s Law : Pressure and Volume
 States that for a given mass of gas at constant
temperature, the volume of a gas varies inversely with
pressure.
 If pressure increases, volume decreases; if pressure
decreases, volume increases.
 P1 x V1 = P2 x V2
P: pressure
V: volume
1: initial condition
2: final condition
 Using Boyle’s Law
1.
A balloon with 30.0L of helium at 103kPa rises to an
altitude where the pressure is only 25.0kPa. What is
the volume of the helium (at constant temperature)?
 Using Boyle’s Law
A balloon with 30.0L of helium at 103kPa rises to an
altitude where the pressure is only 25.0kPa. What is
the volume of the helium (at constant temperature)?
P1=103 kPa V1= 30.0L P2= 25.0kPa V2=?
1.
P1V1= P2V2
V2= P1V1 = (103 kPa)(30.0L) = 124 L
P2
(25.0 kPa)
Since pressure decreases, you expect volume to increase.
2. A gas with a volume of 4.00L at a pressure of 205 kPa is
allowed to expand to a volume of 12.0L. What is the
pressure of the container now (at constant
temperature)?
2. A gas with a volume of 4.00L at a pressure of 205 kPa is
allowed to expand to a volume of 12.0L. What is the
pressure of the container now (at constant
temperature)?
P1= 205 kPa V1= 4.00L
P2=?
V2=12.0L
P1V1=P2V2
P2= P1V1 = (205 kPa)( 4.00 L) = 68.3 kPa
V2
(12.0L)
Classwork: p 121 # 2-6
Charles’s Law: Temperature and Volume
 States that the temperature of an enclosed gas varies
directly with the volume at constant pressure.
 As temperature increases, volume increases.
V1 = V2
T1 T2
V1: initial volume
V2: final volume
T1: initial temperature
T2: final temperature
Temperature has to be in Kelvin scale.
Volume and Temperature
As a gas is heated, it expands.
This causes the density of the
gas to decrease.
23
 Using Charles’s Law
1.
A balloon inflated in a room at 24C has a volume of
4.00L . The balloon is then heated to a temperature
of 58C. What is the new volume ?
 Using Charles’s Law
A balloon inflated in a room at 24C has a volume of
4.00L . The balloon is then heated to a temperature
of 58C. What is the new volume ?
V1 = V2
V1= 4.00L
V2= ?
T1 T2
T1= 24C +273= 297 K
T2= 58C + 273 = 331 K
V2= V1T2 = (4.00L)(331K) = 4.46 L
T1
(297K)
Since temperature increases, you expect the volume to
increase.
Classwork: p124 # 11, 12 (a-c), 13
1.
Combined Gas Law
 Describes the relationship among the pressure,
temperature and volume, when the amount of gas is
constant.
 P1V1 = P2V2
T1
T2
 Standard temperature and pressure (STP): 0C, 1 atm
 Useful conversions:
1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L
Using the combined gas law:
1. The volume of a gas filled balloon is 30.0L at 313K
and 153 kPa. What would the volume be at standard
temperature and pressure (STP)?
Using the combined gas law:
1. The volume of a gas filled balloon is 30.0L at 313K
and 153 kPa. What would the volume be at standard
temperature and pressure (STP)?
P1= 153 kPa V1= 30.0 L T1= 313K
P2= 1 atm=101.3kPa V2= ? T2= 0C= 273K
P1V1 = P2V2
T1
T2
V2= P1V1T2 = (153 kPa x 30.0L x 273K)= 39.5 L
P2T1
(101.3kPa x 313K)
Classwork: p 126 #14(a-c), 15 (a-c) , 16, 17, 19
Ideal Gases
 Objectives
Compute the value of an unknown using the ideal
gas law.
2. Compare and contrast real and ideal gases.
1.
Avogadro’s Principle and Molar Volume
 Avogadro’s principle states that equal volume of all
gases, measured under the same conditions of
pressure and temperature, contain the same number
of particles.
 At STP, the volume of one mole of gas is 22.4 L(this is
called the molar volume)
 From last year, to convert between grams and moles of
a substance we used its molar mass.
Avogadro’s Law
31
Converting between moles and grams:
How many moles are 98.32g CO2?
Calculate molar mass CO2 (use periodic table)
Molar mass= 12 + (2 x 16) =44.0 g/1 mol
To convert grams to moles:
98.32g x 1 mol = 2.23 mol CO2
44g
 Using Avogadro’s Principle
How many moles of carbon monoxide, CO, will
occupy a volume of 250.0 mL at STP?
V= 250.0mL =0.250 L
At STP, molar volume is : 22.4L/1 mol
1.
0.25 L x 1 mol = 0.0112 mol CO
22.4L
Conversions: 1L =1dm3 ; 1 m3 = 1000 L ; 1 cm3 =1mL
2. How many grams of carbon dioxide, CO2, will occupy
a volume of 500.0 mL at STP?
V= 500.0mL =0.5 L
Molar mass CO2= 44g/mol
At STP, molar volume is : 22.4L/1 mol
0.5 L x 1 mol x 44 g = 0.982g CO2
22.4L 1 mol
Classwork: p 132 # 1, 2, 3
The ideal gas law
 Considers that amount of gas varies.
 New variable
 n: number of moles of gas (mol)
 Ideal gas constant (R)
 R= 8.314 L kPa
(when pressure is measured in kPa)
mol K
 R= 0.0821 L atm (when pressure is measured in atm)
mol K
 PV= nRT
(T must be in Kelvin)
Using the ideal gas law
1. A deep underground cavern contains 2.24x106 L of
methane gas (CH4) at a pressure of 1500 kPa and a
temperature of 315K. How many moles of CH4 does
the cavern contain?
Using the ideal gas law
1. A deep underground cavern contains 2.24x106 L of
methane gas (CH4) at a pressure of 1500 kPa and a
temperature of 315K. How many moles of CH4 does
the cavern contain?
P= 1500kPa V= 2.24x106 L R= 8.314 L kPa
n= ?
T= 315 K
mol K
PV=nRT
n= PV = (1500 kPa)x (2.24x106 L )x (mol K)
RT
(8.314 L KPa)x(315K)
=1.28x106 moles
Using the ideal gas law
2. How many moles of oxygen will occupy a volume of
2.5 L at 1.2 atm and 25C? (0.12 moles)
Using the ideal gas law
2. How many moles of oxygen will occupy a volume of
2.5 L at 1.2 atm and 25C? (0.12 moles)
P= 1.2 atm
V=2.5 L T=25C= 298K R= 0.0821 L atm
mol K
Using the ideal gas law
2. How many moles of oxygen will occupy a volume of
2.5 L at 1.2 atm and 25C? (0.123 moles)
P= 1.2 atm
V=2.5 L T=25C= 298K R= 0.0821 L atm
mol K
PV=nRT
n=PV =(1.2atm)x(2.5L)x (mol K)
RT
(0.0821 L atm) x(298K)
= 0.123 moles
Classwork: p 133 # 7,8 and p141 # 3,4
Ideal gas variations
 For calculating molar mass
 M= mRT
PV
M: molar mass (g/mol)
m: mass (g)
 For calculating density
 D= MP
D: density (g/L)
RT
 Sample problem
1.
What is the molar mass of sulfur dioxide, SO2, if 300
mL of the gas has a mass of 0.855 g at STP?
 Sample problem
What is the molar mass of sulfur dioxide, SO2, if 300
mL of the gas has a mass of 0.855 g at STP?
M= ? V= 300mL=0.3 L T=273K, P= 101.3kPa m=0.855g
1.
M= mRT = (0.855g )x(8.314 L kPa)x(273K)
PV
(mol K )
(101.3kPa)(0.3L)
= 63.8g/mol
2. At what temperature will 5.00g of Cl2 exert a
pressure of 115 kPa at a volume of 750 mL?
M= mRT
PV
3. If the density of a gas is 1.2 g/L at 0.920 atm and
20C, what is its molar mass?
D= PM
RT
At what temperature will 5.00g of Cl2 exert a pressure of
115 kPa at a volume of 750 mL?
M= mRT
M= (2x35.5)=71.0g/mol V=750mL=0.750L
PV
T=MPV = 71.0g 115kPa 0.750 L
mol K
= 147 K
mR
5.00 g 8.314 L Kpa
3. If the density of a gas is 1.2 g/L at 0.920 atm and 20C,
what is its molar mass?
D= PM
RT
M= DRT = 1.2g 0.0821 L atm 293 K
=31.4g/mol
P
L
mol K
0.920 atm
2.
Classwork: p 133 #5,6, 9 and p141 #2,5
Gases: Mixtures and Movements
Objectives:
1. Relate the total pressure of a mixture of gases to the
partial pressures of the component gases.
2. Explain how the molar mass of a gas affects the rate
at which the gas diffuses and effuses
 Dalton’s Law
 Partial pressure: the contribution each gas in a
mixture makes to the total pressure.
 Dalton’s law of partial pressure: at constant volume
and temperature, the total pressure exerted by a
mixture of gases is equal to the sum of the partial
pressures of the component of the gases.
 P total= P1 + P2 + P3 + …
 Graham’s Law
 Diffusion: the tendency of molecules to move toward
areas of lower concentration until the concentration is
uniform throughout.
 Effusion: a gas escapes through a tiny hole in its
container.
 Particles of lower molar mass diffuse and effuse faster
than gases of higher molar mass.
 Using Dalton’s and Graham’s Laws:
Air contains oxygen, nitrogen, carbon dioxide, and
trace amounts of other gases. What is the partial
pressure of oxygen (PO2) at 101.3kPa of total pressure
if the partial pressures of nitrogen, carbon dioxide,
and other gases are 79.10kPa, 0.040kPa, and 0.94kPa
respectively?
Ptotal = 101.3 kPa PN2= 79.10kPa PCO2= 0.040kPa
Pothers= 0.94kPa PO2= ?
Ptotal = PN2 + PCO2 + PO2 + Pothers
1.
PO2 = Ptotal- (PN2+ PCO2+ POthers) = 101.3kPa-(79.1kpa+0.040kPa+0.94kPa)
= 21.22 kPa
2. A nitrogen,N2, molecule travels at about 505 m/s at
room temperature. Find the velocity of a helium, He,
atom at the same temperature.
2. A nitrogen,N2, molecule travels at about 505 m/s at
room temperature. Find the velocity of a helium, He,
atom at the same temperature.
rateHe /rateN2=  (molar mass N2/molar massHe)
rateHe = rateN2  (molar mass N2/molar massHe)
= 505m/s (28/4)
= 1336 m/s
Classwork: dalton’s and graham’s law handout.

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