SCH3U-Gases and Atmospheric Chemistry

Unit 5 Ch 11 The Behavoiur of Gases (pg. 417) pg.6

Section Review Pg. 423

1. a) Gas molecules are far apart and can be squeezed together more closely, while liquid

molecules are already very close together, with little room to squeeze them closer together.

b) Gas molecules are far apart as compared to solid molecules. Therefore, there are more solid

particles in a particular volume than gas particles, giving gases a much lower density.

2. An ideal gas is one made of molecules that do not attract each other and that have completely

elastic collisions with anything they hit (ie. another gas molecule or the container wall). They

also do not occupy any space (in science, we say they are point particles, as a point particle has

no dimensions or volume).

3. a) Hexane is made of complete non-polar bonds, it should have relatively low attractive forces.

However, it is made of 21 atoms, and will have some dispersion forces. This molecule would

either be gaseous or liquid at room temperature.

b) Hydrogen fluoride has very low dispersion forces, but is a polar molecule. It is most likely

gaseous at room temperature, but its polar bonds may attract each other strongly enough to

cause it to be a liquid at room temperature.

c) Potassium chloride is an ionic compound. It will be a solid at room temperature.

4. a) As a metal is heated, the metal atoms begin to move and vibrate more quickly, pushing each

other a little further apart, causing the metal to expand. The exact opposite occurs when the

metal is cooled.

b) Since gas molecules do not attract each other, they do not hold each other in place. If no

other object, such as the sides of a container, cause a gas molecule to change its direction

through a collision, gas molecules will continue to move away from each other indefinitely,

causing a gas to have no definite shape or fixed volume.

c) Water molecules in the gas state are very far apart from each other as compared to water

molecules in the liquid state, which stay next to each other. Therefore, the same number of

gaseous water molecules will occupy much more space than the same number of liquid water

molecules.

5. In simple terms in a gas the molecules have maximum movement (translations and rotations

and vibrations), in a liquid they are closer together so restricting their movement (rotations and

vibrations), and in a solid they can only only vibrate, keeping there positions close together.

6. A gas particle will move in a straight line until it collides with something (such as another gas

particle or the wall of a container). At this point, it will change directions (and often speed) and

continue to move in a straight line until another collision occurs.

7. Heating a liquid causes the particles of a liquid to move more quickly. This has the effect of

liquid molecules changing positions with respect to each more quickly. They will also move

slightly further away from each other, causing the volume of the liquid to increase slightly.

Practice Problems p. 434 #1-4

1. V1 = 50.0 cm3 ; V2 = 5.0 cm3 ; P1 = 101.3 kPa

P2 = P1V1

P2 = 101.3 kPa x 50.0 cm3 = 1.0 X 103 kPa

5.0 cm3

2. V1 = 1000 L; P1 = 740.0 torr; P2 = 450.0 torr

V2 = P1V1

V2 = 740.0 torr x 1000 L = 1644 L

450.0 torr

3. V1 = 45.0 cm3 ; P1 = 1.0 atm; P2 = 10.0 atm

V2 = P1V1

V2 = 1.0 atm x 45.0 cm3 = 4.5 cm3

10.0 atm

4. V1 = 45.6 mL; P1 = 490 torr; P2 = 3 atm

P2 = 3 atm x 760 torr = 2 X 103 torr

1 atm

V2 = P1V1

V2 = 490 torr x 45.6 mL = 1 X 10 mL

(2 X 103 torr)

Section Review Pg. 435

1. a) P(kPa) = 2.03 atm X 101.3 kPa = 206 kPa

1.00 atm

b) P(atm) = 85.2 kPa X 1.00 atm = 0.841 atm

101.3 kPa

c) P(torr) = 1.50 atm X 760 torr = 1.14 X 103 torr

1.00 atm

d) P(kPa) = 600 torr X 101.3 kPa = 80.0 kPa

760 torr

2. Air pressure inside a closed container containing gas is created by the gas molecules colliding

with the walls of the container. The number of collisions occurring per second will determine the

magnitude of the pressure. If a closed container’s volume is decreased, the fixed amount of gas

in the container will be squeezed together. The gas molecules inside will be closer together, and

will have a shorter distance to travel in between collisions with the sides of the container. The number of collisions per second will increase, causing the pressure inside the container to

increase.

3. Gas molecules move in straight lines until they hit something, such as the walls of a container

or another moving gas molecules. In a closed container of average size, there are a huge amount

of gas molecules inside the container (recall: one mole of gas contains 6.02 X 1023 particles. Through the sheer number of gas particle collisions occurring per unit time (such as a second),

there will be gas molecules moving in all possible directions. Therefore, gas particle collisions

inside a container will be occurring in all directions, causing pressure to be exerted in all

directions.

4. V1 = 1.00 L; P1 = 102.5 kPa; P2 = 98.6 kPa

V2 = P1V1 ÷ P2

V2 = 102.5 kPa X 1.00 L ÷ 98.6 kPa = 1.04 L

5. V1 = 0.750 L; P1 = 101.3 kPa; V2 = 0.500 L

P2 = P1V1 ÷ V2

P2 = 101.3 kPa X 0.750 L ÷ 0.500 L = 152 kPa

6. V1 = 38.3 mL; V2 = 40.2 mL; P2 = 103 kPa

P1 = P2V2

P1 = 103 kPa x 40.2 mL = 108 kPa

38.3 mL

P1(torr) = 108 kPa x 760 torr = 810 torr

101.3 kPa

The pressure is dropping from the first day to the second day. The weather in the student’s

neighborhood is likely changing from fair weather to rainy or stormy weather.

Practice Problems p. 446 # 5-12

5. a) T(K) = 250C + 273 = 298 K

b) T(K) = 370C + 273 = 310 K

c) T(K) = 1500C + 273 = 423 K

6. a) T(0C) = 373 K - 273 = 1000C

b) T(0C) = 98 K - 273 = -1750C

c) T(0C) = 425 K - 273 = 1520C

7. a) anything at typical late spring or early summer daytime temperature

b) human body

c) food cooking in an oven

8. V1 = 300 mL; T1 = 17°C = 290 K; T2 = 100.0°C = 373.0 K 0 0

V2 = V1T2 ÷ T1 = 300 mL X 373.0 K ÷ 290 K = 386 mL

9. V1 = 2.5 L; T1 = 24.2 C = 297.2 K; T2 = -17.5 C = 255.5 K

V2 = V1T2 ÷ T1 = 2.5 L X 297.2 K ÷ 255.5 K = 2.9 L

10. V1 = 10.0 L; T1 = 20.0 C = 293.0 K; V2 = 30.0 L

T2 = T1V2 ÷ V1 = 293.0 K X 30.0 L ÷ 10.0 L = 879 K = 606°C

11. V1 = 14.5 cm ; T1 = 24.3°C = 297.3 K; V2 = 60 cm3

T2 = V2T1 ÷ V1 = 60 cm3 X 297.3 K ÷ 14.5 cm3 = 1.2 X 103 K = 927°C

Max ∆T= 927°C – 24.3°C= 902°C

12. Step 1: V1 = 600 L; T1 = 25°C = 298 K; T2 = -20° C = 253 K

V2 = V1T2 ÷ T1 = 600 L X 253 K ÷ 298 K = 509 L

Step 2: V1 = 509 L; P1 = 100 kPa; P2 = 400 kPa

V2 = P1V1 ÷ P2 = 100 kPa X 509 L ÷ 400 kPa = 127 L

Practice Problems p. 449

13. P1 = 135.5 kPa; T1 = 15 C = 288 K; T2 = 576 °K

P2 = P1T2 ÷ T1 = 135.5 kPa X 576 K ÷ 288 K = 271 kPa

14. T1 = 18°C = 291 K; P1 = 17.5 atm; T2 = 40°C = 313 K

P2 = P1T2 ÷ T1 = 17.5 atm X 313 K ÷ 291 K = 18.8 atm

15. P1 = 110 kPa; T1 = 45°C = 318 K; P2 = 89 kPa 0

T2 = T1P2 ÷ P1 = 318 K X 89 kPa ÷ 110 kPa = 257 K = -16°C

16. T1 = -7.5°C = 265.5 K; P1 = 206.5 kPa; P2(kPa) = 34.3 psi X 101.3 kPa ÷ 14.7 psi = 236 kPa

T2 = T1P2 ÷ P1 = 265.5 K X 236 kPa ÷ 206.5 kPa = 303 K = 30 °C

Section Review p. 451

1. Compressed gas cylinders should be stored in cool locations and held in place so that the

valves cannot be damaged by the cylinder either falling down or being hit by another object.

High pressure gases will increase their pressure when heated. Since a gas cylinder is designed

to hold gases to a maximum pressure, they should be kept cool so that the gas pressure inside

does not exceed the cylinder’s maximum pressure limit. Also, compressed gases will leave a

cylinder very quickly if the valve is damaged and allows the gas to leave quickly. This can turn

the cylinder into a strong and heavy projectile (there is an episode of MythBusters that proves

gas cylinders can pierce a solid brick wall when their valves are broken off!)

2. a) P1 = 107 kPa; T1 = 300K; T2 = 146 K

P2 = P1T2 ÷ T1 = 107 kPa X 146 K ÷ 300 K = 52.1 kPa

b) V1 = 17 L; T1 = 300 K; T2 = 146 K

V2 = V1T2 ÷ T1 = 17 L X 146 K ÷ 300 K = 8.3 L

3. An equal temperature change (same initial and final temperature, not just temperature change)

causes the same percentage of volume or pressure change. In this case, both the pressure and

volume decreased 48.7% of their inital values.

4. a) When temperature increases, the speed of molecules increases. If the pressure inside a

container is to be kept constant with faster moving molecules, than the volume of the container

must increase so that the frequency of collisions with the walls of the container remains

constant with the faster moving molecules.

b) If the volume of the container remains constant when the speed of the molecules increases,

then the frequency of collisions with the sides of the container will increase, causing an

increase in pressure.

5. As the balloon moves over the hot stove, the heat from the stove increases the temperature of

the gas in the balloon, which causes the speed of the molecules of gas inside the balloon to

increase. Since this increases the frequency of collisions inside the balloon, the flexible walls of

the balloon start to push outwards until the stretching of the balloon causes a tear in the

balloon’s rubber walls, causing it to burst.

Practice Problems p. 457

P1V1 = P2V2

T1 T2

17. V1 = 150 mL; T1 = 260 K; P1 = 92.3 kPa; T2 = 376 K; P2 = 123 kPa

V2 = P1V1T2 = 92.3 kPa X 150 mL X 376 K = 163 mL

P2T1 123 kPa X 260 K

18. P1 = 48 atm; T1 = 290 K; V1 = 0.035 L; V2 = 4.0 L; T2 = 297 K

P2 = P1V1T2 = 48 atm X 0.035 L X 297 K = 0.43 atm

V2T1 4.0 L X 290 K

19. V1 = 48 mL; P1 = 101.3 kPa; T1 = 273 K; V2 = 24 mL; P2 = 110 kPa

T2 = P2V2T1 = 110 kPa X 24 mL X 273 K = 148 K

P1V1 101.3 kPa X 48 mL

20. V1 = 180.0 cm3 ; T1 = 273 K; P1 = 101.3 kPa; V2 = 181.5 cm3 ; P2 = 214.5 kPa 3

T2 = P2V2T1 = 214.5 kPa X 181.5 cm X 273 K = 583 K = 310 0C

P1V1 101.3 kPa X 180.0 cm

21. V1 = 5.0 L; T1 = 478 K; P1 = 350 kPa; V2 = 1.7 L; T2 = 298 K

P2 = P1V1T2 = 350 kPa X 5.0 L X 298 K = 642 kPa

V2T1 1.7 L X 478 K

Practice Problems p. 460

22. Ptot = 98.0 kPa + 202.65 kPa = 300.7 kPa

23. PNe = 116 kPa X 12%/100 = 14 kPa

PHe = 116 kPa X 23%/100 = 27 kPa

PRn = 116 kPa X 65%/100 = 75 kPa

24. Ptot = 325 torr ÷ 40% = 8.1 X 102 torr

25. P(CO2) = 1.00 atm X 70% /100= 0.70 atm

Section Review p. 461

1. V1 = 48 mL; T1 = 273 K; P1 = 101.3 kPa; V2 = 24 mL; P2 = 110 kPa

T2 = P2V2T1 = 110 kPa X 24 mL X 273 K = 148 K

P1V1 (101.3 kPa X 48 mL)

Although the volume of the container is halved, this should cause a temperature increase.

However, the pressure is allowed to increase only a small amount. The temperature of the gas

must have been dropped a great deal to overcome the effect of the smaller volume.

3. The total pressure of a closed system of mixed gases is the result of the individual pressures

exerted by each gas. The ratios of these partial pressures matches the percentage composition

of these gases.

4. The main components of the atmosphere are nitrogen and oxygen. The percentage

composition of nitrogen and oxygen in the atmosphere is fairly constant throughout the entire

atmosphere. Water vapour is also a major component of the atmosphere, but its composition

varies greatly from location to location. Its composition is dependent upon the temperature and

the availability of liquid water in the vicinity.

5. Just before a kernel of popcorn pops, the internal temperature must be at least 100°C, and

perhaps higher. At this time, the pressure inside the kernel is higher than atmospheric pressure.

As the kernel pops, the water vapour escapes and quickly occupies a larger and larger volume.

Its pressure also quickly drops to atmospheric pressure. The temperature of the water vapour

Would then quickly decrease as a result of the larger volume and lower pressure.