PressurePressure: Force applied per unit area.

Barometer: A device that measures atmospheric pressure.

Manometer: A device for measuring the pressure of a gas in a container.

PressureUnits of PressurePascal: (abbrev. Pa) The SI unit for pressure.

1 standard atmosphere = 1.000 atm = 760.0 mm Hg = 760.0 torr

1 standard atmosphere = 101,325 Pa = 101.325 kPa

1.000 atm = 14.69 psi

Pressure and Volume: Boyle’s LawBoyle’s Law: Pressure times Volume equals a constant. PV=k where

k is a constant at a specific temperature for a given amount of gas.

If we know the volume of a gas at a given pressure, we can predict the new volume if the pressure is changed, provided that neither the temperature nor the amount of gas is changed.

Pressure and Volume:Boyle’s LawExampleA sample has a volume of 1.51 L at a

pressure of 635 torr. Calculate the final volume of the gas if the final pressure is 785 torr.

1 1 2 2

1 12

2

(635 )(1.51 ) 1.22(785 )

PV PVPV torr LV LP torr

Volume and Temperature:Charles’s LawCharles’s Law: Proportionality constant times temperature is equal to

volume. V = bT where T is in Kelvins and b is the proportionality constant.

Charles’s Law implies that the amount of gas (moles) and pressure are constant. The volume of the gas is directly proportional to temperature on the Kelvin scale.

Volume and Temperature:Charles’s LawExampleA sample has a temperature of 28oC

and a volume of 23 cm3 at 1 atm. The final temperature was found to be 18oC, assuming no change in pressure. Calculate the final volume.

1 2

1 2

331

2 21

28 273.15 30118 273.15 291

23( ) (291 )( ) 22301

K KK K

V VT T

V cmV T K cmT K

Volume and Moles:Avogadro’s LawAvogadro’s Law: For a gas at constant temperature and pressure,

the volume is directly proportional to the number of moles of gas.

V = an or V = a

n

where V is the volume of the gasN is the number of moles

a is the proportionality constant.

Volume and Moles:Avogadro’s LawExample3H2(g) + N2(g) 2NH3(g)

If one has 15.0 L of H2(g), what volume of N2(g) is required for a complete reaction, given that both gases are at the same temperature and pressure?

22 2

2

1mol N15.0L H 5.00 L N3mol H

The combined gas lawCombined Gas Law: The following equation is called the combined

gas law. It holds when the amount of gas (moles) is held constant.

P1V1 = P2V2

T1 T2

The combined gas lawExampleA sample has a volume of 11.0 L at a

temperature of 13oC and a pressure of 0.747 atm. The sample is heated to 56oC at a final pressure of 1.18 atm. Calculate the final volume.

1 1 2 2

1 2

2

2

(1.18 )(0.747 )(11.0 )(13 273.15 ) (56 273.15 )

8.01

PV PVT T

atm Vatm LK K

V L

Standard (STD) molar volumeLets define the volume occupied by 1 mol of a gas under specified

conditions. For 1 mol of an ideal gas at 273.15 K and 1.0 atm, the volume of the gas is 22.414 L, regardless of gas.

0oC (273.15 K) and 1.0 atm = standard temperature and pressure (STP)

The Ideal Gas LawIdeal gas Law: The equation for the ideal gas law is PV=nRT where

R=0.08206 L atm/mol K (universal gas constant).

Derived from STP and standard molar volume

The Ideal Gas LawExampleA 1.5 mol of a sample of gas has a

volume of 21.0 L at 33oC. What is the pressure of the gas.

(21.0 ) (1.5 )(.08206 )(306 )

1.8

PV nRTL atmP L mol Kmol K

P atm

Density of Gases = m/V PV = nRT n = m/MM PV = (m/MM)RT m/V = PMM/RT =

Dalton’s Law of Partial PressuresPartial Pressure: The pressure that the gas exerts if it were above in

the container. For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present. This can be expressed as Ptotal = P1 + P2 + P3 + . . . . . . where the subscripts refer to the individual gases. The pressures P1, P2, and P3 are the partial pressures.

Important Points1. The volume of the individual gas particles must not be very

important.2. The forces among the particles must not be very important.

Dalton’s Law of Partial PressuresFor a mixture of ideal gases, it is the total number of moles of particles

that is important, not the identity of the individual gas particles. We can calculate the partial pressure of each gas from the ideal gas law.

Dalton’s Law of Partial PressuresExampleA 2.0 L flask contains a mixture of

N2 and O2 gas 25oC. The total pressure of the mixture is 0.91 atm. The mixture is known to contain 0.050 mol of N2. Calculate the partial pressure of O2 and the number of moles of O2 present.

2

2

2

2

2

(2.0 ) (0.050 )(0.08206 )(298 )

0.61

0.91 0.61 0.30

(0.30 )(2.0 ) (0.08206 )(298 )

0.25

N

N

PV nRTL atmP L mol Kmol K

P atm

atm atm atm OPV nRT

L atmatm O L n Kmol Kn molO

Dalton’s Law of Partial Pressures Pressure of gas in mixture of gases =

product of its mole fraction and total pressure of mixture

PA = XAPtot

Ex: Xnitrogen gas = 0.78. What’s its partial pressure at STP?

The Kinetic Molecular Theory of GasesKinetic Molecular Theory: The behavior of individual

particles (atoms or molecules) in a gas.

Postulates of the Kinetic Molecular Theory of Gases1. Gases consist of tiny particles (atoms or molecules).2. These particles are continually in rapid and random

motion.3. The particles are assumed not to attract or to repel each

other.4. All gases, regardless of MM, have = average KE @ =

temp.

The Implications of the Kinetic Molecular TheoryGas speedu2 = (3RT/MM)Maxwell’s equation= gas molecule speeds temperatureSame av. KE @ same tempBUT, different av. Speeds!Smaller MM molecules go faster

The Implications of the Kinetic Molecular Theory Diffusion = random mixing of gases Effusion = gas movt through a small

opening in a container into another container with lower pressure

Graham’s Law: Effusion rate1/effusion rate2 = (MM2/MM1) Gas effusion used initially for 235UF6/238UF6

separation

Gas StoichiometryExample

Calculate the volume of H2 produced at 1.50 atm and 19oC by the reaction of 26.5 g of Zn with excess HCl.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Solution

22

mol Hmol Zn26.5gZn 0.405mol H65.39g mol Zn

PV=nRT

(1.50atm)V=(0.405mol)(0.08206 )(292 )

6.47

L atm Kmol KV L