Chapter 11...Gas Laws The gas laws are simple mathematical relationships between the volume,...
Transcript of Chapter 11...Gas Laws The gas laws are simple mathematical relationships between the volume,...
Chapter 11.2THE GAS LAWS
Background Terminology
Gas Pressure is simply a collision of gas particles with an object. (Pressure is force per unit area)
A vacuum is created (or exists) when there are no (or very few) collisions of gas particles.
Barometers are instruments used to measure atmospheric pressure. Because the particles in air move in every direction, they exert pressure in all directions. This pressure is atmospheric pressure or air pressure.
Air pressure differs at different points on earth because gravity is greater at the surface of the earth. Therefore, air pressure is less at higher altitudes than it is at sea level.
A manometer is an instrument used to measure gas pressure in a closed container.
Units of Pressure
The SI unit of pressure is Pascal (Pa). Pressure can also be reported in the following units:
kilopascal (kPa), millimeters of mercury (mmHg), atmospheres (atm), and Torr
You should know these values and their conversions:
101.3 kPa = 1 atm = 760 Torr = 760 mmHg
Examples:
3.3 atm = ___________ mmHg 564 kPa = __________ atm
745 mmHg = __________ Torr 212 Torr = __________ atm
2508
745 0.279
5.57
Standard Temperature and
Pressure (STP)
Scientists have agreed upon standard gas conditions of
temperature and pressure of a gas. These values are:
Temp: 0˚C Pressure: 1 atm
Temperature conversion: All values of temperature in relation to gas
laws; must be converted to Kelvin.
K = ˚C + 273
Examples: 25˚C = ___________K 313 K = __________ ˚C298 40
Gas Laws
The gas laws are simple mathematical relationships between the volume, temperature, pressure, and the amount of a given gas.
Concepts to remember when performing Gas Law Problems:
Always convert volume to Liters
(1 L = 1000 mL; 1 mL = 1cm3 (cc); 1 L = 1 dm3)
Always convert temperature to Kelvin
( K = ˚C + 273)
Make sure all units of pressure are the same
1 atm = 101.3 kPa = 760 mm Hg = 760 Torr
Boyle’s Law
Relates pressure and volume of a gas
At constant temperature and mass we can predict how a
volume or pressure will change if conditions change.
Boyle’s law states that volume and pressure are inversely
proportional. That is if the volume of a gas increases, the
pressure will decrease.
The mathematical relationship is:
V1 and P1 – represent initial conditions
V2 and P2 – represent final conditionsP
V
V1 P1 = V2 P2
Boyle’s Law Practice
1. A balloon contains 30.0 L of Helium gas at 100.0 kPa. What is the
volume of the balloon when it rises to an altitude where the pressure is only 25 kPa? [Assume constant mass and temp]
2. A sample of oxygen gas has a volume of 150 mL when its pressure
is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?
V1 = 30.0 LP1 = 100.0 kPaV2 = ?
P2 = 25 kPaV1P1 = V2P2 (30)(100) = V2(25) V2 = 120 L
V1 = 150 mLP1 = 0.947 atmV2 = ?P2 = 0.987 atm
V1P1 = V2P2 (0.15)(0.947) = V2(0.987) V2 = 0.144 L
Charles’s Law
Relates the temperature and volume of a gas.
At constant pressure and mass, this law states that
temperature and volume are directly proportional to
each other. Charles’ law states that if the temperature of a gas is increased the volume will also increase.
***Remember temperature must be converted to Kelvin
Mathematical Relationship: V
T
𝑉1
𝑇1
=𝑉2
𝑇2
Charles’s Law Practice
1. A balloon inflated in an air-conditioned room at 27 ˚C has a volume
of 4.0 L. The balloon is then heated to a temperature of 57 ˚C. What is the new volume of the balloon if the pressure remains constant?
2. A gas at 65˚C occupies 4.22 L. At what Celsius temperature will the gas occupy a volume of 3.87 L assuming constant pressure?
V1 = 4.0 LT1 = 27 oCV2 = ?T2 = 57 oC
V2 = 4.4 L
V1 = 4.22 LT1 = 65 oCV2 = 3.87T2 = ?
T2 = 37 oC310 - 273 = T2
𝑉1
𝑇1
=𝑉2
𝑇2
𝑉1
𝑇1
=𝑉2
𝑇2
4.0
300=
𝑉2
330
4.22
338=
3.87
𝑇2
Gay-Lussac’s Law
Gay-Lussac states that the pressure of a gas is directly
proportional to the Kelvin temperature if the volume is
kept constant. In other words if the temperature of a
gas is increased then its pressure will also increase.
The mathematical relationship for this law is: 𝑃1
𝑇1
=𝑃2
𝑇2
Gay-Lussac’s Practice
1. The gas left in a used aerosol can is at a pressure of 100 kPa at
27˚C. If the can is thrown into a fire, what will the internal pressure of the gas be when its temperature reaches 927˚C?
2. A sample of helium gas has a pressure of 1.20 atm at 22˚C. At
what Celsius temperature will the helium reach a pressure of 2.00 atm?
P1 = 100 kPaT1 = 27 oC
P2 = ?T2 = 927 oC
P2 = 400 kPa
P1 = 1.20 atmT1 = 22 oCP2 = 2.00 atmT2 = ? oC
491.667 - 273 = T2
T2 = 219 oC
100
300=
𝑃2
1200
𝑃1
𝑇1
=𝑃2
𝑇2
𝑃1
𝑇1
=𝑃2
𝑇2
1.20
295=
2.00
𝑇2
Combined Gas Law
The three previous gas laws can be combined into a single expression called the combined gas law. This law expresses the relationship between pressure, volume, and temperature of a fixed gas.
The mathematical relationship is
Note: Each formula that we have discussed can be obtained from the combined gas law, if one quantity is held constant.
𝑉1𝑃1
𝑇1
=𝑉2𝑃2
𝑇2
Combined Gas Law
1. The volume of a gas-filled balloon is 30.0 L at 40˚C and 150 kPa pressure. What
volume will the balloon have at STP?
2. A container with an initial volume of 1.0 L is occupied by a gas at a pressure of 150
kPa at 25˚C. By changing the volume, the pressure of the gas increases to 600 kPa
as the temperature is raised to 100˚C. What is the new volume?
P1 = 150 kPaV1 = 30.0 LT1 = 40 oCP2 = 1.00 atmV2 = ?T2 = 0 oC
P1 = 150 kPaV1 = 1.0 LT1 = 25 oCP2 = 600 kPaV2 = ?T2 = 100 oC
V2 = 38.75 L
V2 = 0.313 L
𝑉1𝑃1
𝑇1
=𝑉2𝑃2
𝑇2
𝑉1𝑃1
𝑇1
=𝑉2𝑃2
𝑇2
(150)(30.0)
313=
(𝑉2)(101.3)
273
(150)(1.0)
298=
(𝑉2)(600)
373