C H E M I S T RY 2 0 1 3 - 2 0 1 4

THE BEHAVIOR OF GASES

VARIABLES THAT DESCRIBE A GAS

• Compressibility: a measure of how much the volume of matter decreases under pressure.

• Pressure: a physical force pushing on or against an object; abbreviated P, measured in atmospheres (atm), torr, mmHg, Pascals (Pa), or kilopascals (kPa).• Standard pressure (STP) is 1 atm = 760 torr =

760 mmHg = 101,300 Pa = 101.3 kPa

• Volume: the amount of space an object occupies; abbreviated V, measured in liters, milliliters, cubic meters, or cubic centimeters.

• Temperature: a measurement of the average kinetic energy in an object; abbreviated T, measured in Celsius or Kelvin (use Kelvin for math problems).• Standard

temperature (STP) = 0°C = 273 K

•Mole: a measurement of the number of particles in an object; abbreviated n, measured in moles.

•One mole is equal to 6.02 x 1023 particles (atoms, molecules, or formula units).

THE GAS LAWS

BOYLE’S LAW FOR PRESSURE-VOLUME CHANGES

• For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.

P1V1 = P2V2 or P1 = V2

P2 V1

• You do not need to convert pressure or volume to specific units for these problems because they use ratios.

• A gas is collected in a 242 mL container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at 101.3 kPa? Assume the temperature is constant.

P1V1 = P2V2

P1 = 87.6 kPa V1 = 242 mL

P2 = 101.3 kPa V2 = ?

(87.6)(242) = (101.3)V2

V2= 87.6 * 242 = 209 mL

101.3

CHARLES’ LAW FOR TEMPERATURE-VOLUME CHANGES

• The volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.

V1 = V2 or V1 = T1

T1 T2 V2 T2

• Temperature must be in Kelvin. K = °C + 273

A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm? Assume the pressure is constant.

V1 = V2

T1 T2

V1 = 2.58 L T1 = 15 + 273 = 288 KV2 = ? T2 = 38 + 273 = 311 K 2.58 L = V2___

288 K 311 K

V2 = 2.58 L * 311 K288 K

= 2.79 L

GAY-LUSSAC’S LAW FOR TEMPERATURE-PRESSURE CHANGES

• The pressure of a gas is directly proportional to the Kelvin temperature if the volume is kept constant.

P1 = P2 or P1 = T1

T1 T2 P2 T2

• Temperature must be in Kelvin. K = °C + 273

A 2.0 L flask contains helium gas at a pressure of 685 torr and a temperature of 0°C. If the temperature is raised to 150.°C, what is the new pressure in the flask?

P1 = P2

T1 T2

P1 = 685 torr T1 = 0 + 273 = 273 KP2 = ? T2 = 150. + 273 = 423 K 685 torr =

P2___

273 K 423 KP2 = 685 torr * 423 K 273 K

= 1060 torr

THE COMBINED GAS LAW

• This law combines pressure, volume, and temperature.

• Temperature must be in Kelvin.K = °C + 273

• By canceling out any constant terms, we can derive Boyle’s, Charles’, and Gay-Lussac’s law from the combined gas law.

• Easy way to remember: “Peas and Vegetables on the Table”

• If a helium-filled balloon has a volume of 3.40 L at 25.0ºC and 120.0 kPa, what is its volume at STP?

V1 = 3.40 L

T1 = 25 ºC

P1 = 120 kPa

T2 = 0 ºC

P2 = 101.3 kPa

+ 273 = 298 K

+ 273 = 273 K

V2 = ?

P1V1 P2V2

T1 T2

P1V1 T2

T1 P2

V2

=

=

V2 = (120 kPa) (3.40 L)(298 K )(101.3 kPa)

(273 K )

V2 = 3.69 L

REAL VS. IDEAL GASES

• Ideal gases follow the gas laws at all conditions of temperature and pressure. They do not exist in reality (for example, real gases can be liquefied and sometimes solidified; ideal gases cannot).

• Real gases “deviate” from ideal gas behavior. Real gases behave most ideally at high temperature and low pressure.

THE IDEAL GAS LAW

• The ideal gas law allows us to relate the pressure, volume, number of moles, and temperature of a gas to each other. This law uses the ideal gas constant, R. For the ideal gas law,• Pressure must be in atm• Volume must be in L• n must be moles (it always is)• R = 0.08206 L*atm/mol*K• T must be in Kelvin

A 5.0 L flask contains 0.60 g O2 at a temperature of 22ºC. What is the pressure (in atm) inside the flask? PV = nRTP =

V =

n =

R = 0.0821

T =

5.0 L

1 mol O20.60 g O2

32.0 g O2

= 0.0188 mol O2

?

22ºC

P = nRT V

P =

P =

+ 273

= 295 K

(0.0188 mol) (0.0821) (295 K )

(5.0 L)

0.091 atm

How many grams of krypton are present in a 600. mL container at 1010ºC in which the pressure of krypton is 10.0 atm?

PV = nRT

P =

V =

n =

R = 0.0821

T =

1) find moles of Kr

10.0 atm

600. mL

?

1010 ºC + 273

= 1283 K

Solve for n PV RTn =

n =(10.0 atm) (0.600 L)

(0.0821) (1283 K )

n = 0.0570 mol Kr

= 0.600 L

0.0570 mol Kr 83.8 g Kr1 mol Kr

4.78 g Kr=

0.0570 mol Kr 6.02 x 1023 atoms Kr1 mol Kr

=

3.43 x 1022 atoms Kr

2) Convert to grams

How many atoms is this?

GAS MOLECULES: MIXTURES AND MOVEMENTS

AVOGADRO’S HYPOTHESIS

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

• At STP, 1 mole of particles of any gas = 22.4 L.

DALTON’S LAW OF PARTIAL PRESSURE

• At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures.

Ptotal = P1 + P2 + P3…

Determine the total pressure of a gas mixture that contains nitrogen and oxygen if the partial pressure of the nitrogen is 725 mm Hg and the partial pressure of the oxygen is 426 mm Hg.

Ptotal = P1 + P2 + P3 …

Ptotal = ?

P1 =

P2 =

725 mm Hg

426 mm Hg

Ptotal = 725 mm Hg + 426 mm Hg

Ptotal = 1151 mm Hg

GASES ARE OFTEN COLLECTED BY WATER DISPLACEMENT.

• The total of the gas pressure plus the water vapor pressure is equal to the atmospheric pressure.

Patmospheric = Pgas + PH2O

• When we work a problem like this we must always look up and subtract the water vapor pressure to get the gas pressure.

A sample of N2 gas is collected by the downward displacement of water from an inverted bottle. What is the partial pressure of the N2 gas at 20.0ºC, if the atmospheric pressure is 752 mm Hg? The water vapor pressure is 17.5 mm Hg at 20.0ºC.Ptotal = P1 + P2 + P3 …

Ptotal =

P1 = PN2

752 mm Hg

P2 = PH2O= 17.5 mm Hg

= ?

752 mm Hg = P1 + 17.5 mm Hg

P1 = 734.5 mm Hg

atmospheric P =