• Chapter 14TEST

• Thursday May 17th ,2012; there will also be a Partner lab identification component (w/ some problems).

• Show your work, box your answers for credit

• Some problems require you to know that the following form diatomic molecules:

H2, N2, O2, F2, Cl2, Br2, I2.

• Do The practice problems on the worksheets and review in your book

• Labs: - Charles Law Graph- Molar mass of Unknown Gas- Confirm molar volume of H2 gas- Collect data on properties/reactivity of selected

gases (2 days to collect data)

• Assignments: Mixed Gas Law Worksheet - Vapor pressure worksheet

Gas laws

Charles Law relationship

Gas laws – volume and temperature

Temperature review

• Temperature is a measure of the average kinetic energy in a substance

• Substances at the same temperature have the same average kinetic energy

• As substances heat up, their particles move faster; particles slow down when cooled

• Most substances expand when heated and contract when cooled

• Thermometers contain a liquid which expands and contracts (apx) linearly

• Temperature scales

1. Fahrenheit: water freezes @ 32oF, boils @ 212oFthis scale was invented by Gabriel

Fahrenheit, who was a chemist and glassblower – he made the best early thermometers (ca 1714)

2. Celsius: water freezes @ 0oC, boils @ 100oC

3. Kelvin: water freezes @ 273oK, boils @ 373oK

:same magnitude as Celsius scale

Absolute zero

• All molecular motion stops

• -459 oF, -273 oC, 0 oK

Gas volume and temperature• What happens when a balloon is taken out into the cold?

• Gases linearly expand when heated and contract when cooled

If you take a balloon with a volume of 5570 mL at 21.0 oC outside on a day when the temp is -25.0 oC. Calculate new volume.

Convert C to K

T1 = 21 + 273 = 294 K

T2 = -25 + 273 = 248 K

V2 = (V1)(T2) T1

= (5570mL)(248 K) = 4698 mL294 K

Charles LawV2 = V1

T2 T1

T2 xV2 = V1x T2

T2 T1

V2 = V1x T2

T1

An unopened bag of potato chips initially at 19.0 C has a volume of 259 mL. If left in a closed car during the summer and the temp rises to 58.0 C, what would be its new volume?

V2 = (V1)(T2)/(T1)

= (259 mL)(58 + 273)K / (19 + 273)K

= 294 mL

I make a thermometer which measures temperature by compressing and expanding gas in a piston. At 100 °C the volume of the piston is 20.00 L. If I take the piston outside and the volume is

15.48 L, which activity would be most appropriate?

a) Swimming at the beach

b) rock climbing c) snow skiing

T2 = V2 x T1

V1

= 15.48 L x 373 K

20.00 L T2 = 288.7 K

288.7 K - 273 K = 15.7 C

15.7C x 1.8F/C + 32oF = 60oF

(let’s go rock climbing; too cols for swimming, too hot for snow skiing)

Grahams law of Gas diffusion

• Lighter molecules (at the same temperature) move faster (K.E. = ½ mv2)

• The faster one is (the square root of the ratio of the heavier divided by the lighter mass) times faster ( page 436 of text)

Combined gas lawand ideal gas law

Boyles,Charles, ‘the other one; aka Gay-Lussac’s”

Pressure and temperature• As T ↑ (increases), P ↑ (increases) at a

constant V• Why?• Particles are moving faster, hitting the

walls of the container with greater force and with greater frequency

• And as T ↓ , P ↓ because particles slow down – hit the wall less often and with less force

• P1 = P2 T1 T2 .

Combined gas law

• P1V1 = P2 V2 T1 T2 .

• What would P2 be equal to?

• P2V2 = P1 V1 T2 T1 .

• P2V2(T2) = P1 V1 (T2) T2(V2) T1(V2)

• P2 = P1 V1 (T2) T1(V2)

Combined gas law

• P1V1 = P2 V2 T1 T2 .

Boyles law (at constant temperature)

• P1V1 = P2 V2

Charles law (at constant pressure)

• V1 = V2

T1 T2

Gay Lusac law ( at constant volume)

• P1 = P2

T1 T2

A helium balloon is accidentally released. Its initial volume is 3.95 L , the ground temp is 21.0 oC and pressure is 102 kPa. The balloon rises to 9,900 feet where the temp is 5.00 oC and the pressure is 70.4 kPa. Calculate the volume of the balloon. Remember to convert temp to oK: ( oC + 273 = oK)

V2 = P1 V1 (T2) T1(P2)

= 102 kPa x 3.95 L x 278 K294 K x 70.4 kPa

= 5.41 L

The deepest spot in Lake Harriet is 82 feet. An air bubble with a volume of 2.35 mL rises from this spot. The pressure down there is 337.4 kPa and the temp is 4.0 oC Calculate its volume at the surface where the pressure is 99.23 kPa and the temp is 21.0 oC.

V2 = P1 V1 (T2) T1(P2)

= 337.4 kPa x 2.35 mL x 294 K277 K x 99.23 kPa

= 8.48 mL

Ideal gas law

• 1 mole of any gas has the same volume at the same P and T

• In other words, the same number of molecules of any gas take up the same amount of space under the same conditions

• 1 mole of any gas at STP is 22.4 liters

PV=nRT n is number of molesR is a constant

What is the value of “R”

• P1V1 = P2 V2 = nR T1 T2 .

• (the number of moles times some constant); select the units of pressure and volume you like and calculate “R”

• R = PV/nT

• R= 760 mmHg (22.4 liters)/1 mole(273K)

• R = 62.36 mmHg L / mol. K

What is the value of “R”

• P1V1 = P2 V2 = nR T1 T2 .

• (the number of moles times some constant); select the units of pressure and volume you like and calculate “R”

• R = PV/nT

• R= 101.3 kPa (22.4 liters)/1 mole(273K)

• R = 8.31 kPa L / mol. K

What is the value of “R”

• P1V1 = P2 V2 = nR T1 T2 .

• (the number of moles times some constant); select the units of pressure and volume you like and calculate “R”

• R = PV/nT• R= 1.0 atmosphere (22.4 liters)/1

mole(273K)• R = 0.082 atm. L / mol. K

What is the value of “R”

• R = 62.36 mmHg L / mol. K

• R = 8.31 kPa L / mol. K

• R = 0.082 atm. L / mol. K

• Or any other number and unit combination that equals this value