1

2

Real GasesAn ideal gas adheres to the Kinetic Theory exactly in all situations.Real gases deviate from ideal behavior at high pressures and low temperatures.– When the pressure is high, it becomes more

difficult to compress a gas because the particles actually have a volume of their own.

– When the temperature is low, gas particles slow down and attractions between them become significant as they clump together and form liquids.

3

We must define some terms:

n = moles of gas particlesV = volume (of the container)T = temperature (must be in Kelvin)P = pressure

(You will see these variables in a variety of gas laws)

4

Kelvin is the only temperature scale that measures absolute speed of particles.

K = oC + 273

All temperatures in gas problemsmust be in Kelvin.

Temperature

5

1 atm = 101.325 kPa =760 torr = 760 mmHg

1 atm is the normal atmospheric pressure at sea level.Pressure changes with altitude.Air pressure is measured with a barometer.

Pressure

6

Standard Temperature and Pressure (STP)

Standard Temperature = 0oC = 273 KStandard Pressure = 1 atm or equivalent

7

Pressure, Volume, & Temperature

Boyle’s Law– Pressure and volume are

inversely proportional if the temperature remains constant

P1 V1 = P2 V2Robert Boyle

8

Charles’ Law– Volume and temperature are

directly proportional if pressure remains constant

Temperature must be in Kelvin

Jacques CharlesV1 = V2

T1 T2

9

Gay-Lussac’s Law– Pressure and temperature

are directly proportional if volume remains constant

Temperature must be in Kelvin

Joseph Louis Gay-LussacP1 = P2

T1 T2

Avogadro’s law

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

V = volume of the gas n = number of moles of gas

n1/V1 = n2/V2

Avogadro’s Principle

At STP, 1 mole of gas is equal to 22.4 L

11

12

Combined Gas Law

Temperature must be in KelvinCross out any constants

P1 V1 = P2 V2

T1 T2

13

Sample Problem

At conditions of 785 torr of pressure and 15.0 oC temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745 torr and 30.0oC?

T1 = 15.0 oC + 273 = 288 KT2 = 30.0 oC + 273 = 303 K

14

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

P1=V1= T2 = P2= V2= T1=

785 torr 45.5 mL288 K745 torr?303 K

P1V1T2 = V2

P2T1

(785 torr)(45.5 mL)(288 K)(745 torr)(303 K)

V2 = 50.4 mL

On a cold morning (10.0 oC) a group of hot-air balloonists start filling their balloon with air. After the balloon is three-fourths filled, they turn on the propane burner to heat the air. At what Celsius temperature will the air completely fill the envelope to its maximum capacity of 1700. m3?P1V1 = P2V2

T1 T2

Pressure is constantV1 = ¾ x 1700 m3 V2 = 1700. m3

T1 = 283.15 K T2 = ?

T2 = V2T1

V1

V1T2 = V2T1T2= (1700. m3 )(283 K) 1275 m3

T2 = 377 KT2 = 377 K – 273 = 104 oC

16

Ideal Gas Law

R is the universal gas constant. An “R” value is picked based upon the unit being used to measure pressure.

P V = n R T(R = 0.08206 atm L/mol K)(R = 8.314 kPa L/mol K)

(R = 62.4 mmHg L/mol K)

17

How many moles of a gas at 100.oC does it take to fill a 1.00-L flask to a pressure of 1.50 atm?

n = 0.0490 mol

PV = nRT

P =V = n = R = T =

1.50 atm1.00L?0.0821 Latm/molK373 K

n = PV

RT

n = (1.50 atm)(1.00 L) (0.0821 Latm/molK)(373 K)

18

What is the volume occupied by 9.45 g of C2H2 at STP?

9.45 g C2H2 x 1 mol C2H2 = 0.3629 mol C2H2

26.036 g C2H2

PV = nRT

P =V = n = R = T =

1.00 atm?0.03629 mol0.0821

Latm/molK273 K

V = nRT

P

V = (0.363 mol)(0.0821 Latm/molK)(273 K) 1.00 atm

V = 8.14 L

19

Gas Stoichiometry

Only gas volumes at STP (Avogadro’s Principle 1 mol = 22.4 L) can be entered into a stoichiometry equationIf gas is at a different temperature & pressure, use PV=nRT to convert liters to moles and then continue with stoichiometry

20

3 H2 + N2 2 NH3

A chemist performs this reaction (Haber process) in a chamber at 327oC under a pressure of 900. mm Hg. How many grams of ammonia would be produced from 166.3 L of hydrogen at the above conditions?

PV = nRT n = PV

RTn = (900 mmHg)(166.3 L) (62.4 L mmHg/mol K)(600 K)

n = 4.00 mol H2

4.00 mol H2 x 2 mol NH3 x 17g NH3 = 45.3 g NH3

3 mol H2 1 mol NH3