Reading for Tuesday: Chapter 12 1-4 Homework 11.1 – Due Tuesday 4/14/15

Chapter 11 #'s 2, 4, 6, 10, 12, 13, 14, 18, 20, 24, 26, 30-42 (even)

Homework 11.2 – Due Thursday 4/16/15 Chapter 11 #'s 45, 46-54 (even), 63-66 (all), 68, 74, 76, 80, 82

Lab next Week – EXP 16

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A nitrogen gas sample occupies 50.5 mL at -80.0 ºC and 1250 torr. What is the volume at STP?

1 1250P torr

1 50.5V mL

1 80.0 273 193T K

2 760P torr

2 ?V mL

2 273T K

1 2

22

1

1

V 50.5V 11

TP 127

273

T 19

50

03P 76

m torr

tor

KL

r K mL

1250

76

273

15 .

0 930 5

torr

t mL

K

orr K

1250

7650

0.5

torr

mL

torr50.5 mL

1250 273

1950.5 1

76 317

0

torr

to mL

r

K

Kr mL

1 1 2 2

1 2

P V P V

T T 1

221

1 2TVV

T

P

P

Dalton’s law of partial pressures states that the total pressure of a gaseous mixture is equal to the sum of the individual pressures of each gas.

P1 + P2 + P3 + … = Ptotal

The pressure exerted by each gas in a mixture is its partial pressure, Pn.

An atmospheric sample contains nitrogen, oxygen, argon, and traces of other gases. If the partial pressure of nitrogen is 587 mmHg, oxygen is 158 mmHg, and argon is 7 mmHg, what is the observed pressure as read on the barometer?

2 2P P P Ptotal N O Ar

P 587 158 7total mmHg mmHg mmHg

P 752total mmHg

We can measure the volume of a gas by displacement.

By collecting the gas in a graduated cylinder, we can measure the amount of gas produced.

The gas collected is referred to as “wet” gas since it also contains water vapor.

672.43 mL of hydrogen was collected over water on a day when the barometer read 661.2 mmHg. The temperature of the wet gas was 25 oC.

1) What is the partial pressure of the gas? (vapor pressure of water at 25 oC is 23.8 mmHg)

2 2 2 2P P P P P Ptotal H H O H total H O

2 2P P 661.2 23.P 63 .48 7 mmHgH total H O

672.43 mL of hydrogen was collected over water on a day when the barometer read 661.2 mmHg. The temperature of the wet gas was 25 oC.

2) What is the volume of the gas at STP?

1 637.4 mmHgP

1 672.43V mL

1 25 273 298 KT

2 2P P 661.2 23.P 63 .48 7 mmHgH total H O

2 760P mmHg

2 ?V mL

2 273T K

672.43 mL

11

2

2

12

V 672.43T 273P 637

T 298

.4

P 76V 51

07

K mmHg

mmH

mL

g K mL

P V P-ratio < 1

T V T-ratio < 1

637.4672.43

760

m

mHg

mL

mmHg

637.67

4

7602.4

273

2983

mmHg K mL

mmHg K

637. 273672.43 517

4

76 2980

mmHg

m

mL

m

K

Kg H mL

672.43 mL of hydrogen was collected over water on a day when the barometer read 661.2 mmHg. The temperature of the wet gas was 25 oC.

3) What was the mass of hydrogen collected?

2 2

22

3

2

21 10 1 2.02517 0.0466

1 22.4 1

2

2

H

H

H

HH

H

HH

mol

mL

g

mol

L

g

mL L

An ideal gas is a gas that behaves in a predictable and consistent manner.

Ideal gases have the following properties:gases are made up of very tiny moleculesgas molecules demonstrate rapid motion in straight

lines and in random directionsgas molecules have no attraction for one anothergas molecules undergo elastic collisionsthe average kinetic energy of gas molecules is

proportional to the Kelvin temperature, KE ∝ T

Ideal Gas Law

Ideal Gas Law

1 1 2 2

1 2

P V P V

T T

Avogadro’s Law – Particles and Volume

Particles and volume are directly proportional

1 2

1 2

V V

n n +

Ideal Gas Law

1 1 2 2

1 1 2 2

P V P V

n T n T

almost there!!!!!!

• If we introduce a proportionality constant, R, we can write the equation:

1 1 2 2

1 1 2 2

P V P V

n RT n RT

Ideal Gas Law

1 1 2 2

1 1 2 2

P V P V

n RT n RTPV nRT

pressure (must bein atm)

volume (must bein L)

particles (must bein mol)

temperature (mustbe in K)

Ideal Gas Law

PV nRTatm L

0.08206mol K

R