I. Behavior of

Gases

(Read p. 66-69)

Topic 4- Gases

S.Panzarella

I. Kinetic Molecular Theory – (under ideal circumstances) - explains how an “ideal” gas acts Let’s watch The Kinetic Molecular Theory: Properties of Gases (6:09)…http://education-

portal.com/academy/lesson/the-kinetic-molecular-theory-properties-of-gases.html#lesson

A. Particles in an ideal gas…

• have no volume.

• have elastic collisions.

• are in constant, random, straight-line

motion.

• don’t attract or repel each other.

• have an avg. KE directly related to

Kelvin temperature (KE = ½ MV2) S.Panzarella

B. Gas behavior is most

ideal…

at low pressures

at high temperatures

in nonpolar atoms/molecules of low

molecular mass ie. H2 and He

think about the conditions for a beach vacation

C. Real Gases Particles….

Real gases deviate from

an ideal gas at low

temperatures and high

pressures.

• Have their own volume

(due to increased pressure)

• Attract each other (due to

increased temps

S.Panzarella

Homework

Read page 66, 68-69

pg 4-5 of guide, #’s 1-11

S.Panzarella

Guide pg 4-5 #1-11

1. C

2. C

3. D

4. C

5. B

6. C

7. B

8. Real gases have a small,

but significant volume or

they have weak IMF’s

9. A

10. A

11. A

S.Panzarella

CHEM DO review – You try 5. Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas?

(1) He(g) (2) NH3(g)

(3) Cl2(g) (4) CO2(g)

6. Which gas has properties that are most similar to those of an ideal gas?

(1) N2 (2) O2 (3) He (4) Xe

7. One reason that a real gas deviates from an ideal gas is that the molecules of the real gas have

(1) a straight-line motion

(2) no net loss of energy on collision

(3) a negligible volume

(4) forces of attraction for each other

1. Which gas will most closely resemble an ideal gas at STP? 1) SO2 2) NH3 3) Cl2 4) H2 2. At STP, which gas would most likely behave as an ideal gas? (1) H2 (2) CO2 (3) Cl2 (4) SO2 3. Which gas has properties that are most similar to those of an ideal gas? (1) O2 (2) H2 (3) NH3 (4) HCl 4. Under which conditions does a real gas behave most like an ideal gas? (1) at high temperatures and low pressures (2) at high temperatures and high pressures (3) at low temperatures and low pressures (4) at low temperatures and high pressures

Chem Review Do Answers

1. 4

2. 1

3. 2

4. 1

5. 1

6. 3

7. 4

S.Panzarella

III. The Gas Laws Shows the relationships

between TEMPERATURE, PRESSURE, VOLUME and number of moles of a gas

Used to determine what effect changing one of those variables will have on any of the others.

S.Panzarella

Bill Nye Video

S.Panzarella

Normal conditions…… • Standard Temperature &

Pressure - *STP

• 273 K (or 0◦C) and 101.3 kPa or 1 ATM

• Temperature MUST be (KELVIN) when working with gases.

K = ºC + 273

S.Panzarella

*Found in

Reference Table A

1 Atm = 760 torr =760 mmHg = 101.3 kPa = 14.7 lb/in2

1. Combined Gas law

Combines 3 gas laws

# moles are held constant

P, T and V change Let’s watch (1:05): http://www.youtube.com/watch?v=t-Iz414g

ro&feature=player_embedded

Formula: found on Table T

P1V1

T1 =

P2V2

T2 S.Panzarella

Combined Gas Law – Table T

P1V1

T1 =

P2V2

T2

P1V1T2 = P2V2T1

S.Panzarella

GIVEN:

V1 = 2.00 L

P1 = 80.0 kPa

T1 = 300 K

V2 = 1.00 L

P2 = 240. kPa

T2 = ?

WORK:

P1V1T2 = P2V2T1

(80.0 kPa)(2.00 L)(T2)

= (240. kPa) (1.00 L) (300 K)

T2 = 450 K

COMBINED Gas Law Problem #1

2.00 L sample of gas at 300. K and a pressure of

80.0 kPa is placed into a 1.00 L container at a

pressure of 240. kPa. What is the new

temperature of the gas?

S.Panzarella

GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.3 kPa

T2 = 273 K

WORK:

P1V1T2 = P2V2T1

(71.8 kPa)(7.84 cm3)(273 K)

= (101.3 kPa) V2 (298 K)

V2 = 5.09 cm3

COMBINED Gas Law Problem #2

A gas occupies 7.84 cm3 at 71.8 kPa &

25°C. Find its volume at STP.

S.Panzarella

2. Boyle’s Law

Video: Let’s watch (1:01):

http://www.youtube.com/watch?feature=player

_embedded&v=DcnuQoEy6wA

Pressure vs. Volume

(Constant Temperature):

(think: squeezing a balloon)

As pressure is INCREASED,

volume is DECREASED

S.Panzarella

Boyle’s Law con’t.

INVERSE relationship:

PV = K

formula

P1V1= P2V2

P

V

S.Panzarella

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1T2 = P2V2T1

BOYLE’S Law Problems

A gas occupies 100. mL at 150.

kPa. Find its volume at 200. kPa.

P V

(150.kPa)(100.mL)=(200.kPa)V2

V2 = 75.0 mL

S.Panzarella

3. Charles’ Law Let’s Watch Video (33 sec) : http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded

Volume vs. Temperature (Constant

Pressure) (think: hot air balloon)

As temperature is INCREASED, volume is

INCREASED

• As the temperature of the water increases, the volume of the

balloon increases.

S.Panzarella

Charles’ Law con’t.

DIRECT relationship:

V/T = K

Formula

V1 T2 = V2 T1

V

T

S.Panzarella

GIVEN:

V1 = 5.00 L

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

P1V1T2 = P2V2T1

CHARLES’ Law Problems

A gas occupies 5.00L at 36°C. Find

its volume at 94°C.

T V

(5.00 L)(367 K)=V2(309 K)

V2 = 5.94 L

S.Panzarella

4. Gay-Lussac’s Law

Temperature vs. Pressure (Constant

Volume)- (think: car tires or pressure

cooker)

As temperature is INCREASED,

pressure is INCREASED

S.Panzarella

Gay-Lussac’s Law con’t

DIRECT relationship:

P/T = K

Formula:

P1T2 = P2T1

P

T

S.Panzarella

GIVEN:

P1 = 1.00 atm

T1 = 200 K

P2 = ?

T2 = 800 K

WORK:

P1V1T2 = P2V2T1

GAY-LUSSAC’S Law Problem

A 10.0 L sample of gas in a rigid container at 1.00 atm and 200. K is heated to 800. K. Assuming that the volume remains constant, what is the new pressure of the gas?

P T

(1.00 atm) (800 K) = P2(200 K)

P2 = 4.00 atm

S.Panzarella

Homework

See guide pg 5-6

• Part A (use your RB)

• Part B #12-21

• Quiz on Thursday

S.Panzarella

Topic 4 RB ANSWERS

47) 1

48) 3

49) 1

50) 4

51) 1

52) 3

53) 3

54) 1

55) 2

56) 2

57) 3

58) 3

59) 4

60) 4

61) 3

62) 2

63) 3

S.Panzarella

2 more Gas laws

(TEXT BOOK

p. 350-353)

Read these

pages first!

S.Panzarella

5. Graham’s Law

States that the rate of effusion

(diffusion) of a gas is inversely

proportional to the square root of the gas’s molar mass.

Helium effuses (and diffuses) nearly three

times faster than nitrogen at the same

temperature.

S.Barry S.Panzarella

Graham’s Law cont.

Gases of SMALL MASS (or DENSITY) diffuse faster than gases of higher molar mass.

ex. At STP, which gas will diffuse more rapidly? Use Table S

a) He b) Ar c) Kr d) Xe

S.Barry S.Panzarella

6. Dalton’s Law

At constant volume and temperature,

the total pressure exerted by a mixture of gases

is equal to the sum of the partial pressures of

the component gases.

Formula

Three gases are combined in container T

Ptotal = P1 + P2 + P3 ...

S.Panzarella

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

Dalton’s Law example #1

Hydrogen gas is collected over water at 22.5°C and 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

S.Panzarella

GIVEN:

PO2 = ?

PN2 = 79 kPa

PCO2 = 0.034 kPa

Pothers = 0.95 kPa

Ptotal = 101.3 kPa

WORK:

Ptotal = PO2 + PN2 + PCO2 + Poth

101.3 kPa = PO2 + 79 kPa + 0.034 kPa + 0.95 kPa

101.3 kPa = PO2 + 79.984 kPa

21.316 kPa = PO2

Ex. Air contains oxygen, nitrogen, and carbon dioxide

and other gases. What is the partial pressure of O2

at 101.3 kPa of pressure if the PN2 = 79 kPa , the

PCO2 = 0.034 kPa and Pothers = 0.95 kPa?

Dalton’s Law example #2

S.Panzarella

III. Avogadro’s Hypothesis Let’s watch http://education-portal.com/academy/lesson/molar-volume-using-

avogadros-law-to-calculate-the-quantity-or-volume-of-a-gas.html#lesson

Definition: Equal volume of gases at the same

temperature and pressure contain equal

numbers of particles (molecules) regardless of

their mass

Let’s watch (2:05)

http://www.youtube.com/watch?v=fexEvn0ZOpo&feature=player_embedded S.Panzarella

1) Which rigid cylinder contains the same number of gas molecules at

STP as a 2.0-liter rigid cylinder containing H2(g) at STP?

(a) 1.0-L cylinder of O2(g)

(b) 2.0-L cylinder of CH4(g)

(c) 1.5-L cylinder of NH3(g)

(d) 4.0-L cylinder of He(g)

2) Which two samples of gas at STP contain the same total number of

molecules?

(a) 1 L of CO(g) and 0.5 L of N2(g)

(b) 2 L of CO(g) and 0.5 L of NH3(g)

(c) 1 L of H2(g) and 2 L of Cl2(g)

(d) 2 L of H2(g) and 2 L of Cl2(g)

3) At the same temperature and pressure, which sample contains the

same number of moles of particles as 1 liter of O2(g)?

(a) 1 L Ne(g) (c) 0.5 L SO2(g)

(b) (b) 2 L N2(g) (d) 1 L H2O(ℓ)

B. Molar volume:

The number of molecules in 22.4 L of any gas at STP has been chosen as a standard unit called 1 mole

1 mole = 22.4 L of any gas at STP contains 6.02x1023 particles

22.4 L of any gas at STP is = to its molecular mass

S.Panzarella

Molar volume cont.

Density = molecular mass of gas volume (22.4 L)

ex. What is the density of 1 mole of oxygen gas?

ex. Which gas has a density of 1.70 g/L at STP?

a) F2 b) He c) N2 d) SO2

S.Panzarella

IV. Ideal Gas Law

PV=nRT

Recall - Avogadro’s law, which is

derived from this basic idea, says

that the volume of a gas maintained

at constant temperature and

pressure is directly proportional to

the number of moles of the gas

Video: http://education-portal.com/academy/lesson/the-ideal-gas-law-and-the-

gas-constant.html#lesson

S.Panzarella

PV

T

V

n

PV

nT

Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of-

changes-to-a-gas.html#lesson

= k

UNIVERSAL GAS

CONSTANT

R=0.0821 Latm/molK

R=8.315 dm3kPa/molK

= R

You don’t need to memorize these values!

• Merge the Combined Gas Law with Avogadro’s Principle:

n is the

number of

moles of the

gas

S.Panzarella

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 L

R = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

Ideal Gas Law Problem #1

Calculate the pressure in atmospheres

of 0.412 mol of He at 16°C & occupying

3.25 L.

S.Panzarella

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPa

R = 8.315 dm3kPa/molK

Ideal Gas Law Problem #2

Find the volume of 85 g of O2 at 25°C

and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT

(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3 S.Panzarella