GasesGases

Chapter 13Chapter 13

13.1 The Gas Laws13.1 The Gas Laws

The gas laws apply to ideal gases, which are The gas laws apply to ideal gases, which are described by the kinetic theory in the described by the kinetic theory in the following five statements.following five statements.

Gas particles do not attract or repel each otherGas particles do not attract or repel each other Gas particles are much smaller than the spaces between Gas particles are much smaller than the spaces between

them. them. Gas particles are in constant, random motion.Gas particles are in constant, random motion. No kinetic energy is lost when gas particles collide with No kinetic energy is lost when gas particles collide with

each other or within the walls of their container. each other or within the walls of their container. All gases have the same kinetic energy at a given All gases have the same kinetic energy at a given

temperature.temperature.

Gases are made up of Gases are made up of Two Types of ParticlesTwo Types of Particles

1.1. Molecules Molecules (2 or more atoms bound (2 or more atoms bound together)together)

- Examples: CH- Examples: CH4 4 , O , O2 2 , N, N22

2. Atoms2. Atoms (Noble gases: He, Ne, Ar) (Noble gases: He, Ne, Ar)

Gas particles are widely spread out and Gas particles are widely spread out and moving around:moving around:

Measuring GasesMeasuring Gases

Four variables we use to measure a gas and Four variables we use to measure a gas and talk about its propertiestalk about its properties1.1. AmountAmount

2.2. VolumeVolume

3.3. TemperatureTemperature

4.4. Pressure (force exerted on the walls of its Pressure (force exerted on the walls of its container)container)

Amount of GasAmount of Gas

Quantity of gas is expressed in units of Quantity of gas is expressed in units of MOLESMOLES

MolesMoles allow us to convert to grams of gas, allow us to convert to grams of gas, liters of gas, or the number of gas particles liters of gas, or the number of gas particles we havewe have

Don’t forget the three conversion factors…Don’t forget the three conversion factors… gramsgrams 22.4 L22.4 L 6.02 x 106.02 x 102323 items items

1 mole 1 mole 1 mole 1 1 mole 1 mole mole

VolumeVolume

Volume of a gas = volume of containerVolume of a gas = volume of container Measured in units of Measured in units of LITERSLITERS or or unit of unit of

lengthlength33

1 L = 1000 cm1 L = 1000 cm3 3 = .001 m= .001 m3 3

TemperatureTemperature

The temperature is often measured in The temperature is often measured in degrees Celsiusdegrees Celsius

ALWAYS ALWAYS CONVERT TO KELVIN CONVERT TO KELVIN when when doing calculationsdoing calculations

To Convert:To Convert:

K = °C + 273K = °C + 273

PressurePressure Due to gas particles colliding w/ container (exert Due to gas particles colliding w/ container (exert

force on an area)force on an area) UNITS:UNITS:

– ATMOSPHERES (atm): close to the average ATMOSPHERES (atm): close to the average atmospheric pressure at sea levelatmospheric pressure at sea level What’s atmospheric pressure?What’s atmospheric pressure?

– The pressure on earth’s surface due to the mass of air in the The pressure on earth’s surface due to the mass of air in the atmosphereatmosphere

– Pascal (Pa) or Kilopascal (kPa) Pascal (Pa) or Kilopascal (kPa) 1 atm = 101,325 Pa = 101.3 kPa1 atm = 101,325 Pa = 101.3 kPa

– mm Hg (millimeters of mercury) mm Hg (millimeters of mercury) 1 atm = 760 mm Hg1 atm = 760 mm Hg

– TorrTorr 1 atm = 760 torr1 atm = 760 torr

UNITSUNITS

In other words…In other words…

11 atmatm == 101.3101.3 kPakPa == 101325101325 PaPa = = 760760 mmmm HgHg == 760760 torrtorr

STPSTP

Stands for: Stands for: Standard Temp. and PressureStandard Temp. and Pressure B/c gas behavior depends on B/c gas behavior depends on TemperatureTemperature

and and PressurePressure, we compare properties of , we compare properties of different gases at a standard temperature different gases at a standard temperature and pressureand pressure

STP = 1 atm, 0 ° C STP = 1 atm, 0 ° C

GAS LAWSGAS LAWSThe Mathematical representations The Mathematical representations of relationships between the four of relationships between the four

variables used to describe a gas…variables used to describe a gas…

1. Boyle’s Law (P-V relationship)1. Boyle’s Law (P-V relationship) As pressure DECREASES, volume INCREASESAs pressure DECREASES, volume INCREASES

As pressure INCREASES, volume DECREASESAs pressure INCREASES, volume DECREASES

(inverse relationship)(inverse relationship)

Boyle's Law VideoBoyle's Law Video

Boyle’s Law (Equation)Boyle’s Law (Equation)– AT CONSTANT AT CONSTANT TEMPERATURETEMPERATURE

EQUATION: EQUATION: PP11VV11 = P = P22VV22

Practice Problem (Boyle’s Law)Practice Problem (Boyle’s Law) Suppose I have a 2.5 L balloon filled with He Suppose I have a 2.5 L balloon filled with He

gas. The pressure in the balloon is 1.04 gas. The pressure in the balloon is 1.04 atm. What will the volume of the balloon be atm. What will the volume of the balloon be if I increase the pressure to 1.54 atm?if I increase the pressure to 1.54 atm?

Boyle's Law DEMOS

SolutionSolution

PP11 = 1.04 atm = 1.04 atm PP22 = 1.54 atm = 1.54 atm

VV11 = 2.5 L = 2.5 L VV22 = ? = ?

PP11VV11 = P = P22VV22

(1.04 atm)(2.5 L) = (1.54 atm)(V(1.04 atm)(2.5 L) = (1.54 atm)(V22))

(1.04 atm)(2.5 L)(1.04 atm)(2.5 L) = V = V22

(1.54 atm)(1.54 atm)

1.69 L = V1.69 L = V22

* NOTE: It is important for V to be in LITERS. Convert to L if you are * NOTE: It is important for V to be in LITERS. Convert to L if you are given mL or some other unit of volume.given mL or some other unit of volume.

2. Charles’s Law (T-V Relationship)2. Charles’s Law (T-V Relationship) Consider a gas in a cylinder w/ Consider a gas in a cylinder w/

a moveable piston...a moveable piston...

LOW TEMP

HIGH TEMP

If we DECREASE the temperature, If we DECREASE the temperature, we DECREASE the volume we DECREASE the volume

Heating a Balloon VideoHeating a Balloon Video

Charles’s LawCharles’s Law AT CONSTANT AT CONSTANT PRESSUREPRESSURE……As temperature increases, so does the As temperature increases, so does the

volume of gas.volume of gas.

Charles’s Law: EquationCharles’s Law: Equation

AT CONSTANT AT CONSTANT PRESSUREPRESSURE…… Equation :Equation :

VV11 = V = V22

TT11 T T22

Practice Problem…(You try!)Practice Problem…(You try!)

Say I have a balloon that contains 5.1 L of gas Say I have a balloon that contains 5.1 L of gas at room temperature (25 ° C). What will be the at room temperature (25 ° C). What will be the new volume of the gas if I put it in the freezer new volume of the gas if I put it in the freezer (15° C)?(15° C)?

FIRST convert the temperature values to FIRST convert the temperature values to Kelvin!Kelvin!

K = ° C + 273K = ° C + 273

SolutionSolution VV11 = 5.1 L = 5.1 L VV22 = ? = ?

TT11 = 25 = 25°°C + 273 K = 298 K TC + 273 K = 298 K T22 = 15 = 15°°C + 273 K = 288 KC + 273 K = 288 K

Use the equationUse the equation VV11 = = VV22

TT11 T T22

5.1 L 5.1 L = = VV22

298 K298 K 288 K288 K

(288 K)(5.1 L) (288 K)(5.1 L) = V = V22

298 K298 K

VV2 2 = 4.93 L= 4.93 L

Practice Problem…(You try!)Practice Problem…(You try!)

Say I have a balloon that contains 5.1 L of gas Say I have a balloon that contains 5.1 L of gas at room temperature (25 ° C). What will be the at room temperature (25 ° C). What will be the new volume of the gas if I put it in the freezer new volume of the gas if I put it in the freezer (15° C)?(15° C)?

FIRST convert the temperature values to FIRST convert the temperature values to Kelvin!Kelvin!

K = ° C + 273K = ° C + 273

3. 3. Gay-Lussac’s Law (P-T Relationship)Gay-Lussac’s Law (P-T Relationship)The pressure of a fixed amount of gas varies The pressure of a fixed amount of gas varies directly with temperature when the volume directly with temperature when the volume remains constant. remains constant.

If we INCREASE the temperature, If we INCREASE the temperature, we INCREASE the pressure we INCREASE the pressure

Candle-Water DemoCandle-Water Demo

Egg in a Bottle DemoEgg in a Bottle Demo

Gay-Lussac’s LawGay-Lussac’s Law AT CONSTANT AT CONSTANT VOLUMEVOLUME……As temperature increases, so does the As temperature increases, so does the

pressure of gas.pressure of gas.

Gay-Lussac’s Law: EquationGay-Lussac’s Law: Equation

AT CONSTANT AT CONSTANT VOLUMEVOLUME…… Equation :Equation :

PP11 = P = P22

TT11 TT22

Practice Problem…(You try!)Practice Problem…(You try!)

The pressure of oxygen gas inside a canister is 5.00 The pressure of oxygen gas inside a canister is 5.00 atm at 25.0°C. The canister is located at a camp high atm at 25.0°C. The canister is located at a camp high on Mount Everest. If the temperature there falls to -on Mount Everest. If the temperature there falls to -10.0°C, what is the new pressure inside the canister?10.0°C, what is the new pressure inside the canister?

REMEMBER convert the temperature values REMEMBER convert the temperature values to Kelvin!to Kelvin!

K = ° C + 273K = ° C + 273

SolutionSolutionPP11 = P = P22

TT11 TT22

PP11 = 5.00 atm, T = 5.00 atm, T11 = 25.0°C + 273 = 298 K = 25.0°C + 273 = 298 K

PP22 = ????, T = ????, T22 = -10.0°C + 273 = 263 K = -10.0°C + 273 = 263 K

5.00 atm5.00 atm = = PP2 2

298 K 263 K298 K 263 K

PP2 2 = 4.41 atm= 4.41 atm

4. 4. Avogadro’s Principle Avogadro’s Principle (V-n Relationship)(V-n Relationship)

Equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

Avogadro’s Avogadro’s LawLaw

AT CONSTANT AT CONSTANT TEMP and PRESSURETEMP and PRESSURE……The volume of a gas is completely dependent The volume of a gas is completely dependent

on how many moles of gas are in the on how many moles of gas are in the container.container.

Avogadro’s Law: EquationAvogadro’s Law: Equation

AT CONSTANT AT CONSTANT TEMP and PRESSURETEMP and PRESSURE…… Equation :Equation :

VV11 = V = V22

nn11 nn22

((n= number of molesn= number of moles))

Remember 1 mol = 22.4 L at STPRemember 1 mol = 22.4 L at STP

Practice Problem…(You try!)Practice Problem…(You try!)

How many moles of acetylene (CHow many moles of acetylene (C22HH22) gas occupy ) gas occupy

a volume of 3.25 L at STP?a volume of 3.25 L at STP?

SolutionSolutionVV11 = V = V22

nn11 nn22

VV11 = 3.25 L, n = 3.25 L, n11 = ???? mol = ???? mol

VV22 = 22.4 L, n = 22.4 L, n22 = 1 mol = 1 mol

3.25 L3.25 L = = 22.4L22.4L

nn11 1 mol 1 mol

nn1 1 = 0.145 mol= 0.145 mol

5. Dalton’s Law of Partial 5. Dalton’s Law of Partial Pressures…Pressures…

In a mixture of gases, each gas exerts a In a mixture of gases, each gas exerts a pressure as if it were on its ownpressure as if it were on its own

The total pressure of a mixture of gases is The total pressure of a mixture of gases is the sum of the pressure due to each gasthe sum of the pressure due to each gas

PT = P1 + P2 + P3 + …

The total pressure (PT) is the sum of the pressures from each gas (P1 = pressure of gas 1, P2 = pressure of gas 2, etc)

Practice ProblemPractice Problem

What is the atmospheric pressure if the What is the atmospheric pressure if the partial pressures of nitrogen, oxygen, and partial pressures of nitrogen, oxygen, and argon (the components of air) are 604.5 mm argon (the components of air) are 604.5 mm Hg, 162.8 mm Hg, and 0.5 mm Hg, Hg, 162.8 mm Hg, and 0.5 mm Hg, respectively?respectively?

SolutionSolution

Use Dalton’s Law…Use Dalton’s Law… PPTT = P1 + P2 + P3 + … = P1 + P2 + P3 + …

P = 604.5 + 162.8 + 0.5 P = 604.5 + 162.8 + 0.5 P = 767.8 mm HgP = 767.8 mm Hg

Gas Law SummaryGas Law SummaryLaw Equations Constant

Boyle’s P1V1= P2V2 Temperature & amount

Charles’s V1= V2

T1 T2

Pressure & amount

Gay-Lussac’s P1= P2

T1 T2

Volume & amount

Avogadro’s V1= V2

n1 n2

Temperature & pressure

Dalton’s Partial Pressure

PT= P1 + P2 + P3

Putting the Laws Together…Putting the Laws Together…

Combined gas law:Combined gas law:PP11VV11 == PP22VV22

nn11TT11 nn22TT22

The Ideal Gas LawThe Ideal Gas Law

Combining the Laws into one Combining the Laws into one equation…equation…

THE IDEAL GAS LAW: THE IDEAL GAS LAW:

- Describes the behavior of an ideal gas in Describes the behavior of an ideal gas in terms of P, T, and nterms of P, T, and n

- Ideal Gas:Ideal Gas: gas described by kinetic gas described by kinetic molecular theorymolecular theory

- Real gases behave like ideal gases under Real gases behave like ideal gases under ordinary conditions ordinary conditions (doesn’t apply to (doesn’t apply to VERY LOW T and VERY HIGH P)VERY LOW T and VERY HIGH P)

The Value of “R”The Value of “R” R is called the “gas constant”R is called the “gas constant” The value of R depends on the units you The value of R depends on the units you

need to use to solve a problem need to use to solve a problem

Value of RValue of R UnitsUnits

0.08210.0821 atm-Latm-Lmole-Kmole-K

8.3148.314 Pa-mPa-m33

mole –Kmole –K

Units:Units:

Which value of “R” do I use??Which value of “R” do I use?? Depends on the units of P and VDepends on the units of P and V P can be in atm or PaP can be in atm or Pa V can be in L or mV can be in L or m33

T is ALWAYS in KT is ALWAYS in K

Practice Problem 1: PV = nRTPractice Problem 1: PV = nRT How many moles of gas are How many moles of gas are

contained in a 2.0 L container with contained in a 2.0 L container with a pressure of 1.5 atm at 100°C?a pressure of 1.5 atm at 100°C?

Write out the variables given. Write out the variables given. Convert T to Kelvin if needed.Convert T to Kelvin if needed.

P = P = V =V = T = T = What variable is missing?What variable is missing?

Which value of R do you need to Which value of R do you need to use?use?

Value of RValue of R UnitsUnits

0.08210.0821 atm-Latm-Lmole-Kmole-K

8.3148.314 Pa-mPa-m33

mole –Kmole –K

8.314 8.314 J____J____mole- K mole- K

Practice Problem 1: PV = nRTPractice Problem 1: PV = nRT How many moles of gas are How many moles of gas are

contained in a 2.0 L container with a contained in a 2.0 L container with a pressure of 1.5 atm at 100°C?pressure of 1.5 atm at 100°C?

Write out the variables given. Convert Write out the variables given. Convert T to Kelvin if needed.T to Kelvin if needed.

P = 1.5 atmP = 1.5 atm V = 2.0 LV = 2.0 L T = 100°C + 273 = 373 KT = 100°C + 273 = 373 K What variable is missing?What variable is missing? nn Which value of R do you need to use?Which value of R do you need to use? 0.0821 0.0821 atm-Latm-L

mole-Kmole-K

Value of RValue of R UnitsUnits

0.08210.0821 atm-Latm-Lmole-Kmole-K

8.3148.314 Pa-mPa-m33

mole –Kmole –K

8.314 8.314 J____J____mole- K mole- K

Practice Problem 1:Practice Problem 1:

Plug the values into the equation and solve Plug the values into the equation and solve for the unknown variable…for the unknown variable…

PV = PV = nRTnRT

(1.5 atm)(2.0L) = n (0.0821(1.5 atm)(2.0L) = n (0.0821atm-L/mole-K)(373 K)atm-L/mole-K)(373 K)

(1.5 atm)(2.0L)__________(1.5 atm)(2.0L)__________ = 0.098 moles = 0.098 moles (0.0821(0.0821atm-L/mole-K)(373 K)atm-L/mole-K)(373 K)

If the gas is carbon dioxide, how If the gas is carbon dioxide, how many grams of gas are produced?many grams of gas are produced?

Carbon dioxide = COCarbon dioxide = CO22

0.098 moles x 0.098 moles x 44 grams44 grams = 4.312 grams = 4.312 grams

1 mole1 mole