Chapter 11Chapter 11

Molecular Molecular Composition of GasesComposition of Gases

Avogadro’s LawAvogadro’s LawEqual Volumes of Gases at Equal Volumes of Gases at the Same Temperature & the Same Temperature & Pressure contain the Pressure contain the Same Same Number of “Particles.”Number of “Particles.”

BalloonsBalloons Holding 1.0Holding 1.0 L L

of Gas at 25ºof Gas at 25º C and C and

1 atm.1 atm.

Each balloon containsEach balloon contains

0.0410.041 mole of gas ormole of gas or

2.5 x 102.5 x 102222 molecules molecules..

Avogadro’s LawAvogadro’s LawFor a gas at constant temperature For a gas at constant temperature and pressure, the volume is directly and pressure, the volume is directly proportional to the number of proportional to the number of moles of gas (at low pressures).moles of gas (at low pressures).

VV = = anan

aa = proportionality constant = proportionality constant

VV = volume of the gas = volume of the gas

nn = number of moles of gas = number of moles of gas

What are the four quantities needed to describe a gas?

• Pressure• Volume• Temperature• Moles of gas

In Chapter 10, which three of these quantities did we vary?

• Pressure• Volume• Temperature

Now we will consider the moles of gas.

Moles of GasMoles of Gas

The number of moles of gas will The number of moles of gas will always affect at least one of the always affect at least one of the other three quantities.other three quantities.

Ideal Gas LawIdeal Gas Law Mathematical relationship Mathematical relationship

among among

pressure,pressure, volume, volume, temperature andtemperature and the number of moles.the number of moles.

Ideal Gas LawIdeal Gas Law Mathematical equationMathematical equation

coming from the combination coming from the combination ofof

- Boyle’s Law- Boyle’s Law

- Charles’ Law &- Charles’ Law &

- Avogadro’s Law- Avogadro’s Law

Ideal Gas LawIdeal Gas Law An An equation of state equation of state for a gas.for a gas. ““state” is the condition of the gas at state” is the condition of the gas at

a given time.a given time.

PVPV = = nRTnRT

An Ideal Gas is a hypothetical substance.

Ideal Gas Law is an empirical equation.

Ideal Gas LawIdeal Gas Law

PPVV = = nnRRTT

RR = proportionality constant = proportionality constant

= ideal gas constant = ideal gas constant

= 0.0821 L atm = 0.0821 L atm mol mol

See Table 11-1 on page 342 for additional values for R.

Ideal Gas LawIdeal Gas LawPPVV = = nnRRTT

Because of the units of R,Because of the units of R,

PP = pressure in atm = pressure in atm

VV = volume in liters = volume in liters

nn = moles = moles

TT = temperature in Kelvins = temperature in Kelvins

Holds closely at Holds closely at PP < 1 atm < 1 atm

StandardStandard Temperature Temperature

and Pressureand Pressure““STP”STP”

PP = 1 atmosphere = 1 atmosphere

TT = = CC

Ideal Gas LawIdeal Gas LawPPVV = = nnRRTT

If gas is at STP,If gas is at STP, RR = 0.0821 L atm = 0.0821 L atm mol mol

PP = 1.00 atm = 1.00 atm TT = 273 Kelvin = 273 Kelvin Then Then VV /n = volume in liters/ mole /n = volume in liters/ mole = RT/ P= RT/ P

= 0.0821 = 0.0821 L atmL atm x 273 K x 273 K mole Kmole K

1.00 atm1.00 atm

= 22.4 L/ mole= 22.4 L/ mole

Standard Molar Volume of a GasStandard Molar Volume of a Gas

The volume occupied by one mole of The volume occupied by one mole of a gas at STP.a gas at STP.

22.4 L/ mole22.4 L/ mole

A Mole of Any Gas Occupies a A Mole of Any Gas Occupies a

VolumeVolume of Approximately 22.42 L of Approximately 22.42 L

at STPat STP

ProblemProblem

What is the volume of 77.0 g of What is the volume of 77.0 g of nitrogen dioxide at STP?nitrogen dioxide at STP?

SolutionSolution77.0 g NO77.0 g NO22 x x 1 mole NO1 mole NO22 x x 22.4 L22.4 L

46.01 g NO46.01 g NO22 1 mole 1 mole

= 37. 5 L NO= 37. 5 L NO22

ProblemProblem

What is the mass of 1.33 x 10What is the mass of 1.33 x 1044 mL of oxygen gas at STP?mL of oxygen gas at STP?

SolutionSolution

1.33 x 101.33 x 1044 mL x mL x 1 L1 L x x 1 mol1 mol

1000 mL 22.4 L1000 mL 22.4 L

x x 32.00 g O32.00 g O22

1 mole O1 mole O22

= 19.0 g O= 19.0 g O22

ProblemProblemAt STP, 3 L of chlorine is produced during a At STP, 3 L of chlorine is produced during a chemical reaction. chemical reaction.

What is the mass of this gas?What is the mass of this gas?

SolutionSolution 3 L x 3 L x 1 mole 1 mole x x 70.90 g Cl70.90 g Cl22

22.4 L 1 mole Cl22.4 L 1 mole Cl22

= 9 g Cl= 9 g Cl22

ProblemProblem

What pressure, in atmospheres, is What pressure, in atmospheres, is exerted by 0.325 mole of hydrogen exerted by 0.325 mole of hydrogen gas in a 4.08 L container at 35gas in a 4.08 L container at 35 C? C?

NotNot at STP!!! at STP!!!

Use Ideal Gas Equation!!!Use Ideal Gas Equation!!!

Solution: Solution: P V = n R TP V = n R T

P = nRT/VP = nRT/V

T (K) = 35 + 273 = 308 KT (K) = 35 + 273 = 308 K

P = 0.325 mole x 0.0821 P = 0.325 mole x 0.0821 L atm L atm x 308 K x 308 K

mole Kmole K

4.08 L4.08 L

P = 2.01 atmP = 2.01 atm

ProblemProblemA sample that contains 4.38 mole of a gas A sample that contains 4.38 mole of a gas at 250 K has a pressure of 0.857 atm. at 250 K has a pressure of 0.857 atm. What is the volume? What is the volume?

Solution: PV = nRTSolution: PV = nRT

V = nRT/PV = nRT/P

V = 4.38 mole x V = 4.38 mole x 0.0821 L atm 0.0821 L atm x 250 K x 250 K

mole Kmole K

0.857 atm0.857 atm

V = 105 LV = 105 L

Molar Mass of a GasMolar Mass of a Gas

n = # of molesn = # of moles

= = grams of gasgrams of gas = = (m) (m)

molar massmolar mass molar mass molar mass

P V = n R T

P V = (m/molar mass) R T

Therefore,molar mass = m R T

P V

ProblemProblemAt 28At 28C and 0.974 atm, 1.00 L of gas has a C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g.mass of 5.16 g.

What is the molar mass of this gas?What is the molar mass of this gas?

SolutionSolution

molar mass = m R T P V

Molar Mass = 5.16 g x 0.0821 L atm x 301 K mole K

0.974 atm x 1.00 L

Molar Mass = 130.92 g/mole = 131 g/mole

Also, since density (d) = m

VThen, since

molar mass = m R T P V

molar mass = d R T P

Density = (molar mass) x P R T

Temperature in K!!

Gas density usually measured in g/L

ProblemProblem What is the density of a sample of What is the density of a sample of ammonia gas, NHammonia gas, NH33, if the pressure is , if the pressure is 0.928 atm and the temperature is 0.928 atm and the temperature is 63.0 63.0 C?C?

SolutionSolution

Density = (molar mass) x P R T

Density = 17.04 g/mole x 0.928 atm0.0821 L atm x 336 K

mole K

Density = 0.573 g/ L NH3

ProblemProblemThe density of dry The density of dry air at sea levelair at sea level

(1 atm) is 1.225 g/L (1 atm) is 1.225 g/L at 15at 15C.C.

What is the What is the average molar average molar mass of the air?mass of the air?

SolutionSolution

Density = (molar mass) x P R T

Therefore,

Molar Mass = Density x R x T P

SolutionSolution

Molar Mass = Density x R x T P

Molar Mass = 1.225 Molar Mass = 1.225 g g x 0.0821 x 0.0821 L atmL atm x 288 K x 288 K

L mole KL mole K

______

1 atm1 atm

Molar Mass = 28.96 g/mole = 29.0 g/moleMolar Mass = 28.96 g/mole = 29.0 g/mole

HomeworkHomeworkRead pages 338-339.Read pages 338-339.

Complete Worksheet.Complete Worksheet.