UNIT 2 REVIEW

UNIT 2 REVIEW

Part 1

(Pages 181–182) 1. A

2. C 3. B 4. 310 5. 121 6. 3.1 7. D 8. B 9. 1, 3, 4, 2 10. C 11. B 12. 1.2 13. 2, 2, 4, 1 14. A 15. 12.4 16. D 17. B Solutions

4. T = (37 + 273) K = 310 K 5. P

V

= P

V

V

=

2

PV P

= 102 kPa 250 mL 210 kPa

u = 121 mL

or V

= 102 kPa

250 mL u

210 kPa = 121 mL

According to Boyle’s law, the final volume of air will be 121 mL.

6. T

= (22 + 273) K = 295 K T

= (–15 + 273) K = 258 K

1 1

V T =

2

V T V

=

1

T V T

= 258 K 3.5 L 295 K

u = 3.1 L

or V =

258 K 3.5 L

295 K

u = 3.1 L

According to Charles’ law, the final volume of the balloon will be 3.1 L.

12. T

= (–23 + 273) K = 250 K

T

= (12 + 273) K = 285 K

1 1 1

PV

T =

2

P V T V =

1 2

PV T T P

= 102 kPa u 1.00 kL u 285 K 250 K u 96 kPa = 1.2 kL

or V

= 285 K 102 kPa

1.00 kL

250 K

u u

96 kPa = 1.2 kL

According to the combined gas law, the final volume of air will be 1.2 kL.

15. 2 NH

NO

(s) Æ 2 N

(g) + 4 H

O(g) + O

(g) 49.6 L V

O2

V = 1

49.6 L

u 4 = 12.4 L or

O2

V = 49.6 L H O

2

1 L O 4 L H O

u = 12.4 L

According to the law of combining volumes, the volume of oxygen produced will be 12.4 L.

Part 2

(Pages 182–185)

18. (a) T = (0 + 273) K = 273 K (b) T = (21 + 273) K = 294 K (c) T = (–273 + 273) K = 0 K 19. (a) 4.00 atm 101.325 kPa

1 atm

u = 4.05 10 kPa = 0.405 MPa u

(b) 763 mm Hg 101.325 kPa

760 mm Hg

u = 102 kPa

(c) 450 atm 101.325 kPa 1 atm

u = 4.56 10 kPa = 45.6 MPa u

20. (a) _n

_{5.1 L}

L 1 mol

u 24.8 0.21 mol (b) n

20.7 m L

L 1 mol 22.4

u 0.924 mol

[Note: These problems can also be solved using the ideal gas law.]

21. (a)

H2

500 mol

V 24.8 L

1 mol

u 12.4 kL

(b)

H S2

56 k mol

V 24.8 L

1 mol

u 1.4 ML

22. (a) The volume of a gas sample decreases proportionally with its pressure, assuming the amount and temperature remain constant.

(b) The volume of a gas sample increases proportionally with its absolute temperature, assuming the amount and pressure remain constant.

(c) The volume of a gas sample is directly proportional to the product of its amount and its absolute temperature, and is inversely proportional to its pressure.

23. A law is empirical. For example, the volume–temperature relationship is a law. A theory is based on non-observable ideas such as the random, colliding motion of molecules.

24. Avogadro’s idea is theoretical because it is based on a non-observable concept. Molecules in a gas cannot be seen or counted directly.

25. (a) At low temperatures and high pressures, the behaviours of ideal and real gases differ the most. The volume of an ideal gas will approach zero; the volume of a real gas will reach a constant, non-zero value as the gas condenses to a liquid. According to the kinetic molecular theory, the particles of an ideal gas have negligible size and no forces exist between the molecules. However, a real gas has molecules with a definite size and attractive intermolecular forces. As the molecules slow down (at low temperatures) and become closer together (at high pressure), the intermolecular forces become significant and the molecules bond together to form a liquid.

(b) At high temperatures and low pressures, real gases behave very much like ideal gases.

According to the kinetic molecular theory, the molecules are moving at high speeds (at high temperature) and are relatively far apart (at low pressure). Any effect of actual molecular size and intermolecular forces will be minimal. Under these conditions the assumptions of the kinetic molecular theory for ideal gases will be valid.

26. P

V

= P

V

V

=

2

PV P

= 100 kPa 28.8 L 350 kPa

u = 8.23 L

or V

= 100 kPa 28.8 L u

350 kPa V

= 8.23 L

According to Boyle’s law, the final volume of hydrogen gas will be 8.23 L, assuming the chemical amount and temperature are constant.

27.

1 2

1 2

2 1 2

1

(23 273) K 296 K

3.5 L 296 K

259 K 4.0 L

T

V V

T T

T V T V



u t

= (259 – 273) °C = -14 °C

According to Charles’s law, the outside temperature is -14 °C.

28. T = (140 + 273) K = 423 K PV = nRT

Br2

n = PV RT

= 60 kPa 18.8 L 8.314 kPa

u

• L 413 K

mol • K u

= 0.33 mol

Br2

1 mol • K

or 18.8 L

8.314 kPa

n u 60 kPa

• L u 0.33 mol

413 K

According to the ideal gas law, the chemical amount of bromine is 0.33 mol.

29. T

= (25 + 273) K = 298 K T

= (–15 + 273) K = 258 K

1 1 1

PV

T =

2

P V T V

=

1 2

T PV T P

= 258 K 100 kPa u 5.00 L 298 K 91.5 kPa

u u

= 4.73 L

or V

= 100 kPa

5.00 L u

91.5 kPa

258 K 298 K u

V

= 4.73 L

According to the combined gas law, the final volume of helium in the balloon will be 4.73 L.

30. Boyle’s Experiment Problem

What is the effect of pressure on the volume of a gas?

Design

A volume of air is placed inside a syringe. The pressure on the air inside the syringe is changed by adding different weights onto the end of the syringe plunger. The weight added is the manipulated variable and the volume measured is the responding variable. Two important controlled variables are temperature and chemical amount of air.

Charles’ Experiment Problem

What is the effect of temperature on the volume of a gas?

Design

A volume of air is placed inside a syringe that is then immersed in a water bath. The temperature of the water bath is manipulated and the volume of air is measured as the

responding variable. Two important controlled variables are pressure and chemical amount of air.

31. (a) 180 kPa 1 atm 101.326 kPa

u 1.78 atm

(b)

2 1

(15 273) K 288 K (40 273) K 313 K T

T P V





1 2

1

P V T

2 2 1 2 2

1

180 kPa 313 K

196 kPa 288 K

T P PT

T

u

According to the combined gas law, the tire pressure becomes 196 kPa.

(c) According to the kinetic molecular theory, an increase in temperature means an increase in the average speed of the molecules. Because the volume is constant, the molecules will collide with each other and the walls of the tire more often and with more force,

therefore, the pressure increases.

(d) If the tire pressure is set when the tires are very hot, the pressure may become very low when the tires cool to ambient temperatures. It should be noted that significant under- inflation may damage tires.

32. T

= (19.5 + 273) K = 292.5 K P V

1

P V

T T

T

=

1

T P P

= 292.5 K 195 kPa u 96.7 kPa = 590 K

t

= (590 – 273) °C = 317 °C

or T

= 195 kPa

292.5 K u

96.7 kPa = 590 K

t

= (590 – 273) °C = 317 °C

According to the combined gas law, the final gas temperature when the container breaks will be 317 °C.

33. (a) T

= (150 + 273) K = 423 K T

= (110 + 273) K = 383 K

1 1 1

PV

T =

2

P V T P

=

1 2

T PV TV

= 383 K 600 kPa 10.0 kL u u 423 K 18.0 kL u = 302 kPa

or P

= 10.0 kL 600 kPa u

18.0 kL

383 K

u 423 K

= 302 kPa

According to the combined gas law, the final pressure of the turbine steam will be 302 kPa.

(b) PV = nRT

H O2

n PV

RT

600 kPa 10.0 kL 8.314 kPa

u

• L 423 K mol • K

1.71 kmol

u

H O2

1.71 k mol

m 18.02 g

1 mol

u 30.7 kg

H O2

1 mol

or m 10.0 k L u • K

8.314 kPa

600 kPa

• L u 18.02 g

423 K u 1 mol 30.7 kg According to the ideal gas law, the mass of steam is 30.7 kg.

34. According to the kinetic molecular theory, warm air has faster-moving molecules that occupy a larger volume at a constant pressure compared with cooler air. The volume increase makes such air less dense than surrounding air, so it rises, as denser, cooler air around it falls.

35. (a) 4 NH

(g) + 5 O

(g) Æ 4 NO(g) + 6 H

O(g) V V 1.00 L

All gas volumes measured at the same temperature and pressure.

NH3

V = 4

1.00 L

u 4 = 1.00 L

O2

V = 5

1.00 L

u 4 = 1.25 L

or V

= 1.00 L NO 4 L NH

4 L NO

u = 1.00 L

O2

V = 1.00 L NO 5 L O

4 L NO

u = 1.25 L

According to the law of combining volumes, the required volume of ammonia is 1.00 L and of oxygen is 1.25 L.

(b) Avogadro’s theory states that equal volumes of gases at the same temperature and

pressure contain equal numbers of molecules. Therefore, the ratio of coefficients (number of molecules) in the equation is the same as the ratio of volumes measured.

36. (a) When 50 mL of oxygen reacts with excess dissolved glucose, the balanced chemical equation indicates that the same volume (chemical amount) of carbon dioxide gas will be produced because they both have the same coefficients in the balanced chemical

equation. Therefore, 50 mL of carbon dioxide gas will be produced.

(b) The first reaction will produce greater leavening. The leavening depends on the chemical amount of carbon dioxide produced. More carbon dioxide is produced per mole of glucose consumed in the first reaction compared to the second reaction. In the first reaction, three times more carbon dioxide is produced than in the second reaction (6 mol versus 2 mol).

37.

16.05 g 1 mol

d 1 mol

u 0.647 g/L

24.8 L

N2

28.02 g 1 mol

d 1 mol

u 1.13 g/L

24.8 L

You should be near the floor because methane is less dense (0.647 g/L) than nitrogen (1.13 g/L), which is the largest component of air.

38. (a) According to Boyle’s law, the pressure and volume are inversely related. The new volume will be:

300 mL u 1

2 = 150 mL.

(b) The solubility of carbon dioxide gas in the beverage is also important.

(c) The pop is carbonated because there is a large pressure of carbon dioxide above the liquid, which keeps most of the gas dissolved in the pop. When the can is opened, there is a sudden decrease in pressure reducing the solubility and causing bubbles to form so rapidly that some of the liquid is carried out of the container.

39. Purpose

The purpose of this investigation is to create a possible relationship between two variables.

Problem

What effect does the pressure of nitrogen gas have on its solubility in water at a fixed temperature?

(a) Analysis

According to the evidence of the graph, the solubility of nitrogen in water increases with increased pressure.

(b) From the graph, approximately 1.5 mmol/L of nitrogen will dissolve at 225 kPa.

Therefore, in 5.00 L of blood (mostly water), the chemical amount of nitrogen that will dissolve is:

N2

5.00 L

n 1.5 mmol

u 1 L 7.5 mmol

(c) From the graph, at 100 kPa the solubility of nitrogen gas is 0.65 mmol/L.

The difference in the solubility of nitrogen from 300 kPa to 100 kPa is (2.00 – 0.65) mmol/L = 1.35 mmol/L at 25 °C.

2

1.35 mmol

5.00 L 6.75 mmol

N

1 L

n u

The chemical amount of nitrogen that comes out of solution is 6.75 mmol.

PV = nRT

N2

V = nRT P

=

6.75 m mol 8.314 kPa

u • L

mol 298 K

• K 100 kPa

u

= 167 mL

According to the evidence and the ideal gas law, the volume of nitrogen gas that will come out of solution is 167 mL.

(d) Scuba diving involves using compressed gases (usually air) for breathing under water. The process of breathing under water also requires that the pressure of the air be the same as the external water pressure in order to properly and safely expand and contract the lungs. Finally, the solubility of air (nitrogen) decreases as the pressure decreases, causing nitrogen to come out of solution. If a diver ascends too quickly, nitrogen comes out of solution and forms bubbles in the blood vessels, causing a painful condition known as “the bends.” Therefore, knowledge of gas properties is important for scuba diving.

40. Purpose

The purpose of this investigation is to use gas concepts in a chemical analysis.

Problem

Is the HCFC sample tested CHF

Cl(g), C

H

FCl

(g), or C

H

F

Cl(g)?

Analysis

m = 457.64 g – 454.26 g = 3.38 g T = (22.0 + 273) K = 295 K PV = m

M RT

§ ·

¨ ¸

© ¹

M = mRT PV

=

3.38 g 8.314 u kPa • L

mol • K u 295 K 100.1 kPa u 0.840 L = 98.6 g/mol

or M 3.38 g u 8.314 kPa • L mol • K

295 K u 100.1 kPa

1 0.840 L

u 98.6 g/mol

According to the ideal gas law, the molar mass of the gas is 98.6 g/mol. The substance tested must be C

H

F

Cl(g) (M = 100.50 g/mol), which has the closest molar mass to the

experimental value.

41. Problem

What is the relationship between the volume of a sealed balloon and the temperature to which it is subjected?

Prediction

As the temperature of the water surrounding the balloon increases, the volume of the balloon

will also increase.

Design

An inflated balloon is tied (sealed) and then placed in a series of water baths of varying temperatures. After the balloon has been submerged in each water bath for 3 min, the circumference of the balloon (measured consistently along the same line of the balloon) is measured. The results are then compiled, compared, and interpreted.

Materials Ɣ balloon Ɣ pail

Ɣ water of varying temperatures Ɣ outdoor alcohol thermometer Ɣ waterproof marker

Ɣ tape measure Procedure

1. Inflate a balloon to about half of its maximum volume and tie it.

2. Draw a line around the centre of the balloon (circumference) about which you will measure. (Use a waterproof marker).

3. Measure the circumference of the balloon.

4. Record the room temperature.

5. Fill a pail with enough cold water to completely submerge the balloon.

6. Measure and record the temperature of the water in the pail.

7. Totally submerge the balloon in water for at least 3 min.

8. Remove the balloon and immediately measure its circumference.

9. Gradually increase the temperature of the water in the pail and repeat steps 5 to 8 for as wide a range of temperatures as possible.

Analysis

Ɣ Use the circumference to calculate the volume of the balloon after each trial (assuming that the balloon is spherical).

Ɣ Convert the temperatures in Celsius to Kelvin.

Ɣ Graph the volume of the balloon versus the absolute temperature.

42. Normally, the air temperature decreases with altitude near Earth’s surface. Combustion products from automobiles and industrial processes usually rise as part of a convection cell.

As these combustion products rise, the polluted air is carried away from ground level, and fresh air comes in to replace it. When a temperature inversion occurs, warmer air at higher altitudes moves on top of the air near the ground. Convection can no longer occur. Under these conditions, the polluted air is unable to rise through the warmer air aloft, so it remains trapped under a layer of warm air.

43. Ɣ Gases assume the volume and shape of a container, flow easily, and are very

compressible. These familiar observations can easily be explained by the concepts of the kinetic molecular theory. The familiar observation that hot air rises can be explained by a combination of kinetic molecular theory and the concepts of molar volume and molar mass (related to atomic theory). The observation that many gases can be condensed to liquids is related to the concept of real versus ideal gases and intermolecular forces.

Ɣ Pressure, temperature, volume and chemical amount are all related by the ideal gas law, PV = nRT.

Extension

44. The earliest gases used as anaesthetics were nitrous oxide, ether, and chloroform.

Unfortunately, some of these gases were unreliable and some were flammable. Cyclopropane

was later discovered to be an anaesthetic but (like ether) it was flammable. By the 1950s,

continued research and development led to other non-flammable and safer anaesthetics like halothane mixed with oxygen or air. Halothane and related compounds are still used today.

Compressed air is the common gas for shallow dives. There are many different mixtures that are used for deep dives. For example, heliox is a mixture of helium and oxygen. Trimix is a mixture of oxygen, nitrogen, and helium that is used for very deep dives. The percentage composition used depends on the depth of the dive. Other noble gases are also occasionally used in mixtures for breathing or as a decompression gas.

45. Bernoulli proposed that a gas contained tiny particles (“corpuscles”) that moved very rapidly.

These particles collide with the walls of the container creating the gas pressure. This idea was not accepted because scientists at the time believed that particles of a gas repelled each other and stayed more or less in one place in a fluid called the ether. The accepted belief was supported by other work from people like Isaac Newton.

The lack of acceptance of Bernoulli’s hypothesis illustrates several things about the nature of science. It is difficult to get new ideas accepted unless there is some experimental evidence to clearly contradict the existing belief and support the new idea. Scientists tend to stick to accepted beliefs until they are forced to change. At the time, the new hypothesis did not provide any better explanation than the accepted beliefs which fit with other ideas at the time. Finally, this may also illustrate that well-known and established scientists have a greater influence than relatively unknown scientists.

46. (a) A large amount of methane rising through the ocean would create a large area of bubbles.

This mixture would not be dense enough to float a ship. Vessels caught in such an area would sink abruptly.

(b) Methane rising through the atmosphere would create an area of gas much less dense than air. Airplanes caught in such an area would lose lift, drop in altitude abruptly, and crash if the drop took them down to the surface of the water.

(c) One popular hypothesis for the supposed anomalies of the Bermuda triangle is that the

disappearances were due to aliens from outer space. The problem with this or any other

explanation is that evidence clearly shows that some disappearances were in fact due to

severe weather and this region is no more dangerous than any other part of the ocean.