CH110 Chpt 7 Gases CH110 Chapter 8: Gases Kinetic Molecular Theory Pressure Gas Laws.
Gas Laws: Introduction At the conclusion of our time together, you should be able to: 1. List 5...
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Transcript of Gas Laws: Introduction At the conclusion of our time together, you should be able to: 1. List 5...
Gas Laws: IntroductionAt the conclusion of our time together,
you should be able to:
1. List 5 properties of gases2. Identify the various parts of the kinetic
molecular theory3. Define pressure4. Convert pressure into 3 different units5. Define temperature6. Convert a temperature to Kelvin
Importance of Gases
• Airbags fill with N2 gas in an accident.
• Gas is generated by the decomposition of sodium azide, NaN3.
• 2 NaN3 ---> 2 Na + 3 N2
General Properties of Gases
• There is a lot of “free” space in a gas.
• Gases can be expanded infinitely.• Gases fill containers uniformly and
completely.• Gases diffuse and mix rapidly.
To Review
Gases expand to fill their containers
Gases are fluid – they flow Gases have low density
1/1000 the density of the equivalent liquid or solid
Gases are compressible Gases effuse and diffuse
Properties of Gases
Gas properties can be modeled using math. This model depends on —
• V = volume of the gas (L)• T = temperature (K)
– ALL temperatures in the entire unit MUST be in Kelvin!!! No Exceptions!
• n = amount (moles)• P = pressure
(atmospheres)
Ideal Gases
Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases consist of tiny particles that
are far apart relative to their size. Collisions between gas particles and between particles and the walls of the container are elastic collisions
No kinetic energy is lost in elastic collisions
Ideal Gases (continued)
Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion
There are no forces of attraction between gas particles
The average kinetic energy of gas particles
depends on temperature, not on the identity of the particle.
Pressure
Is caused by the collisions of molecules with the walls of a container
Is equal to force/unit area SI units = Newton/meter2 = 1 Pascal (Pa) 1 atmosphere = 101.325 kPa (kilopascal)1 atmosphere = 1 atm = 760 mm Hg = 760
torr1 atm = 29.92 in Hg = 14.7 psi = 0.987 bar
= 10 m column of water.
Measuring Pressure
The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century.The device was called a “barometer”
Baro = weight Meter = measure
An Early Barometer
The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.
PressureColumn height measures
Pressure of atmosphere• 1 standard atmosphere (atm)
*= 760 mm Hg (or torr) *= 29.92 inches Hg *= 14.7 pounds/in2 (psi)= 101.325 kPa (SI
unit is PASCAL)
Let’s Review:Standard Temperature and
Pressure“STP”
Allows us to compare amounts of different gases by converting them all to STP (the same temperature and pressure conditions) P = 1 atmosphere, 760 torr T = 0°C, 273 Kelvin The molar volume of an ideal gas is
22.42 liters at STP
Pressure Conversions
A. What is 475 mm Hg expressed in atm?
B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg?
= 1520 mm Hg
= 0.625 atm475 mm Hg x
29.4 psi x
1 atm760 mm Hg
760 mm Hg 14.7 psi
101.325 kPa 14.7 psi
Pressure Conversions – Your Turn
A. What is 2.00 atm expressed in torr?
B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?
= 1520 torr2.00 atm x 760 torr 1 atm
= 221 kPa32.0 psi x
Converting Celsius to Kelvin
Gas law problems involving temperature require that the temperature be in KELVINS!
Kelvins = C + 273
°C = Kelvins - 273
Don’t use temp. when determining sig. figs. for your answer
Charles’ Law
If n and P are constant, then V α T
V and T are directly proportional.
• If one temperature goes up, the volume goes up!
Jacques Charles (1746-1823). Isolated
boron and studied gases. Balloonist.
1 2
1 2
V V
T T
Gay-Lussac’s Law
If n and V are constant, then P α T
P and T are directly proportional.
• If one temperature goes up, the pressure goes up!
Joseph Louis Gay-Lussac (1778-1850)
1 2
1 2
P P
T T
Boyle’s Law
P α 1/VThis means Pressure and
Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.
Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.
1 1 2 2PV PV
Boyle’s Law ExampleBoyle’s Law Example
A bicycle pump is a good example of Boyle’s law.
As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.
Combined Gas Law (for a fixed amount, moles, of gas)
• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! (if a variable is not given in the problem, just leave it out)
1 1 2 2
1 2
PV PV
T T
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = 0.800 atm V1 = 180 mL T1 = 302 K
P2 = 3.20 atm V2 = 90 mL T2 = ??
Calculation
• P1 = 0.800 atm V1 = 180 mL T1 = 302 K
• P2 = 3.20 atm V2 = 90 mL T2 = ??
P1 V1 P2 V2 = T1 T2
Solving for T2 = 604 K
T2 = 604 K - 273 = 331 °C
Gas Laws: Avogadro’s and Ideal
At the conclusion of our time together, you should be able to:
1. Describe Avogadro’s Law with a formula.2. Use Avogadro’s Law to determine either
moles or volume3. Describe the Ideal Gas Law with a formula.4. Use the Ideal Gas Law to determine either
moles, pressure, temperature or volume5. Explain the Kinetic Molecular Theory
Avogadro’s Law
Equal volumes of gases at the same T and P have the same number of molecules.
V and n are directly related.
twice as many molecules
Avogadro’s Law Summary
For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).
V1
n1
V2
n2
Standard Molar Volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro
Ultimate Comined Gas Law(include Aavogadro’s law in the combined gas law)
1 1 2 2
1 1 1 2
PV PV
T Tn n
The relationship between T, P, n, and V for a gas is a constant which we call the ideal gas constant, R (=0.08206 L atm/ mole K). So we can can set one side as equal to R (the gas constant) and it is now called the ideal gas law.
IDEAL GAS LAW
Brings together gas properties.
Can be derived from experiment and theory.
BE SURE YOU KNOW THIS EQUATION!
P V = n R T
Ideal Gas Law
PV = nRT P = pressure in atm (you may need to
convert)
V = volume in liters n = moles R = proportionality constant
= 0.08206 L atm/ mol·K T = temperature in Kelvin
R is a constant, called the Ideal Gas ConstantInstead of learning a different value for R for all the
possible unit combinations, we can just memorize one value and convert the units to match R.
R = 0.08206
L • atm mol • K
Using PV = nRT
How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?
Solution1. Get all data into proper units
V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg)
= 0.98 atmAnd we always know R, 0.08206 L atm / mol K
How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
Solution2. Now plug in those values and solve for the
unknown. PV = nRT
RT RT
Using PV = nRT
n = (0.98 atm)(2.7 x 10 4 L)
(0.0821 L • atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
Gas Laws: Dalton’s LawAt the conclusion of our time together,
you should be able to:
1. Explain Dalton’s Law 2. Use Dalton’s Law to solve a problem
Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
This is particularly useful in calculating the pressure of gases collected over water.
Dalton’s Law of Partial Pressures
What is the total pressure in the flask?Ptotal in gas mixture = PA + PB + ...
Therefore, Ptotal = PH2O + PO2
= 0.48 atm
Dalton’s Law: total P is sum of PARTIAL pressures.
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
0.32 atm 0.16 atm
Gases in the Air
The % of gases in air Partial pressure (STP)
78.08% N2 593.4 mm Hg
20.95% O2 159.2 mm Hg
0.94% Ar 7.1 mm Hg
0.03% CO2 0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg 2 2 2
Total Pressure = 760 mm Hg
Collecting a gas “over water”
• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?
768 torr – 17.5 torr = 750.5 torr
Dalton’s Law of Partial Pressures
Also, for a mixture of gases in a container, because P, V and n are directly proportional if the other gas law variables are kept constant:
nTotal = n1 + n2 + n3 + . . .
VTotal = V1 + V2 + V3 + . . .
This is useful in solving problems with differing numbers of moles or volumes of the gases that are mixed together.
Gas Laws: GasStoichiometry
At the conclusion of our time together, you should be able to:
1. Use the Ideal Gas Law to solve a gas stoichiometry problem.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Solution1.1 g H2O2 1 mol H2O2 1 mol O2 22.4 L O2
34 g H2O2 2 mol H2O2 1 mol O2
= 0.36 L O2 at STP
Gas Stoichiometry: Practice!
How many grams of He are present in 8.0 L of gas at STP?
= 1.4 g He
8.0 L He x 1 mol He
22.4 L He
x 4.00 g He
1 mol He
Gas Stoichiometry Trick
If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.
3 H2(g) + N2(g) 2NH3(g)
3 moles H2 + 1 mole N2 2 moles NH3 67.2 liters H2 + 22.4 liter N2 44.8 liters NH3
Gas Stoichiometry Trick Example
How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature?
3 H2(g) + N2(g) 2NH3(g)
12 L H2
L H2
= L NH3 L NH3
3
28.0
Gas Stoichiometry Example on HO
How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate?
2 KClO3(s) 2 KCl(s) + 3 O2(g)
10.5 g KClO3 1 mol KClO3
122.55 g KClO3
3 mol O2
2 mol KClO3
0.13 mol O2
Gas Stoichiometry Example on HO
How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate?
2 KClO3(s) 2 KCl(s) + 3 O2(g)
3.2 L O2
(1.0 atm)(V) (0.08206 atm*L/mol*K) (298 K)(0.13 mol)
Gas Laws: Dalton, Density and Gas Stoichiometry
At the conclusion of our time together, you should be able to:
1. Explain Dalton’s Law and use it to solve a problem.
2. Use the Ideal Gas Law to solve a gas density problem.
3. Use the Ideal Gas Law to solve a gas stoichiometry problem.
Gas Stoichiometry #4
nRT
VP
How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s) 2 KCl(s) + 3 O2(g)
= “n” mol O2
50.0 g KClO3 1 mol KClO3
122.55 g KClO3
3 mol O2
2 mol KClO3 = 0.612 mol O2
L atm(0.612mol)(0.0821 )(310K)
mol K0.930atm
= 16.7 L
Try this one!
How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
Gas Laws: Finding “R”At the conclusion of our time together, you
should be able to:
1. Use the Ideal Gas Law to find the value of “R”.
See the problem on p. 50
Partial Pressure of CO2
Carbon Dioxide gas over water at 17.0 degrees C and 97.932 kPa. What is the pressure of the CO2 gas?
97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm
1.90 kPa
Determine the Moles of CO2 Used:
Mass of canister before-mass of canister after
17.099 g – 16.524 g= 0.575 g0.575 g CO2 1 mol CO2
44.01 g CO2
0.0131 mol CO2