Chapter 12 (Practice Test) The Gas Laws

d. Temperature T (Kelvin) 1. Identify the four factors that affect gases. a. Pressure P b. Volume V c. Number of moles n d. Temperature T (Kelvin)

Dalton’s Law of Partial Pressure (F) 2. PTotal = P1 + P2 + P3 + . . . Pn P1V1 P2V2 Combined Gas Law (D) 3. = T1 T2 Boyle’s Law (A) 4. P1V1 P2V2 = Ideal Gas Law (E) 5. PV = n RT Graham’s Law (G) 6. or V1 V2 Charles’s Law (B) 7. = T1 T2 P1 P2 Gay-Lussac’s Law (C) 8. = T1 T2

Identify STP: Standard temperature _____________________ Standard pressure ________________________ 0°C 273 K 101.3 kPa 760 mmHg 1 atm.

At what point during the process of cooling do the gas laws become ineffective? When the gas condenses into a liquid. Gas laws don’t work for substances that are not gases. Gas laws deviate from ideal behavior at high pressures and low temperatures.

Use the Kinetic Theory to describe why a decrease in the volume of a gas will cause and increase in pressure as stated by Boyle’s Law. Particles moving at the same speed (same temp.) in a smaller space will hit the walls more often. Since pressure is the result of collisions with the walls, the pressure will increase with more collisions.

PV (a) Boyle’s Law: P1V1 = P2V2 (c) Graham’s Law: or P1 V1 P2 V2 Which law shows an inverse relationship between pressure and volume? 13. Which law compares the rates of two gases? 14. Identify the Gas Law used if the pressure was held constant. 15. What would happen to the pressure exerted by a gas if the volume is decreased and temperature is held constant? (a) Boyle’s Law: P1V1 = P2V2 (c) Graham’s Law: or P1 V1 P2 V2 (b) Charles’s Law: = T1 T2 (a) Pressure would increase. PV PV

PV (c) Kelvin Pressure would decrease not increase. 16. In order to solve problems using Charles’s Law or the Combined Gas Law the temperature must be measured in: Which of these changes would NOT cause an increase in the pressure of a gaseous system? The container is made larger. Additional amounts of the same gas are added to the container. The temperature is increase. Another gas is added to the container. (c) Kelvin PV PV Pressure would decrease not increase.

(c) The temperature decreases. This is how they make liquid air. 18. What happens to the temperature of a gas when it is expanded quickly? 19. What happens to the pressure of a gas inside a container if the temperature of the gas is lowered? 20. As the temperature of the gas in a balloon decreases ____. (c) The temperature decreases. This is how they make liquid air. (Joule-Thomson expansion engine) P1 P2 (c) The pressure decreases. = T1 T2 Direct relationship so… P and V should decrease also. P1 P2 V1 V2 = = T1 T2 T1 T2 Temp. measures KEave (b) Lower temp. = lower KEave.

P V (a) Pressure increases. or 21. If a balloon is squeezed, what happens to the air pressure within the balloon. 22. Which of the following molecules would have the greatest velocity if each gas had the same kinetic energy? a. Br2 b. Cl2 c. NH3 d. H2 e. Ar P V (a) Pressure increases. 35.45 x 2 70.90 14.01 x 1 1.01 x 3 + 3.03 = 17.04 1.01 x 2 2.02 79.90 x 2 159.80 39.95 or The smallest molecule travels the fastest.

Chapter 12 (Practice Test) The Gas Laws (Problem Section)

P1V1 P2V2 = V2 P2 P2 V1 = P1 = V2 = P2 = V2 = 300.0 L 55.0 kPa A gas sample has a volume of 300.0 L when under a pressure of 55.0 kPa. What is the new volume if the pressure is increased to 165 kPa while the temperature is held constant. V1 = 300.0 L P1V1 P2V2 = V2 P1 = 55.0 kPa P2 P2 V2 = ______ L Divide both sides by P2 P2 = (55.0 kPa) (300.0 L) 165 kPa V2 = 165 kPa Calculator 55 x 300 ÷ 165 = 100 L V2 = 100. L or 1.00 x 102 L Answer w/ 3 Sig. Figs.

V1 V2 = T1 T2 = T2 Solve for T2 so cross multiply then divide V1 = 2. A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the volume of the gas be 402 mL, assuming there is no change in pressure? V1 = 804 mL V1 V2 = T1 = 300. K 27 °C + 273 = T1 T2 Convert to KELVINS! 804 (402) V2 = = 402 mL 804 (300) 804 T2 ______ K T2 = Calculator 402 x 300 ÷ 804 = 150 K Solve for T2 so cross multiply then divide T2 = 150. K or 1.50 x 102 K Answer w/ 3 Sig. Figs.

P1 P2 = T1 T2 = = = 100. kPa or 1.00 x 102 kPa P1 = 3 x 102 kPa 3. The gas in a closed container has a pressure of 3.00 x 102 kPa at 30C. What will the pressure be if the temperature is lowered to –172.0C ? P1 = 3 x 102 kPa = 300. kPa P1 P2 = T1 = 303. K 30 °C + 273 = T1 T2 Convert to KELVINS! P2 = Solve for P2 ______ kPa (101) 300 P2 P2 (300) 303 (101) (101) T2 = = 101.0 K -172.0 °C + 273 = 101 303 = = 100. kPa or 1.00 x 102 kPa Answer w/ 3 Sig. Figs. Calculator 101 x 300 ÷ 303 = 100. kPa

P1 V1 P2 V2 = T1 T2 = = = 100 L or 1 x 102 L V1 = 100 L P1 = 80 kPa 4. A 100 L sample of gas is at a pressure of 80 kPa and a temperature of 200 K. What volume does the same sample of gas occupy at STP ? V1 = 100 L P1 V1 P2 V2 = P1 = 80 kPa T1 T2 T1 = 200 K Solve for V2 V2 = ______ L (100) (200) (80) (273) (101.3) (80) V2 (273) (100) 101.3 V2 (273) = P2 = 101.3 kPa 273 (101.3) (101.3) 200 T2 = 273 K = = 100 L or 1 x 102 L Answer w/ 1 Sig. Fig. Calculator 273 x 80 x 100 ÷ 101.3 ÷ 200 = 107.79862 L

P V = n R T = = = P P P P = 1,400 kPa or 1.4 x 103 kPa P = ____ V = 5. What is the pressure of a 5.0 L container which contains 3.0 moles of a gas at a temperature of 0C? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K) 5. What is the pressure of a 5.0 L container which contains 3.0 moles of a gas at a temperature of 0C? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K) P V = n R T P = ____ V = 5.0 L P (5) = (3) (8.31) (273) n = 3.0 mol Solve for P R = 8.31 kPa•L mol•K kPa P P (5) = = (3) (8.31) (273) (5) (5) T = 0°C = 273 K P = 1,400 kPa or 1.4 x 103 kPa Answer w/ 2 Sig. Figs. Calculator 3 x 8.31 x 273 ÷ 5 = 1,361.178 kPa

H P V = n R T = = = V = 150 L = 1.5 x 102 L P = 120 kPa V = ____ (120) 6. What volume does 14.0 g of Hydrogen gas (H2) take up at a temperature of 30C and a pressure of 120 kPa? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K) 6. What volume does 14.0 g of Hydrogen gas (H2) take up at a temperature of 30C and a pressure of 120 kPa? Note: Use Ideal gas law. R=8.31 (kPa x L)/(mol x K) P V = n R T P = 120 kPa V = ____ (120) V = (6.93) (8.31) (303) n = 1 mol H2 14.0 g H2 x = 6.93 mol Solve for V 2.02 g H2 1 H Hydrogen 1.01 1.01 x 2 2.02 = = (120) V (6.93) (8.31) (303) R = 8.31 kPa•L mol•K L (120) (120) T = 30°C + 273 = 303 K V = 150 L = 1.5 x 102 L Answer w/ 2 Sig. Figs. Calculator 6.93 x 8.31 x 303 ÷ 120 = 145.4104575 L

= = = 2 CH4 is 2 times faster than SO2 Lighter gases travel faster 7. Two gases CH4 and SO2 are released at the same time from opposite ends of the room. You are in the center of the room. Which gas will reach you first? (hint: mass of S=32; C=12; O=16; and H=1.0) Part A: Circle one: Methane (CH4) or Sulfur dioxide (SO2) Part B: How much quicker will the gas reach you? 12 x 1 12 + 1 x 4 4 = 16 32 x 1 32 + 16 x 2 32 = 64 Lighter gases travel faster v1 = CH4 or m1 = 16 g/mol v2 = SO2 CH4 64 4 = = = 2 m2 = 64 g/mol SO2 16 1 CH4 is 2 times faster than SO2

= = PT PT ____ kPa PT = 300 kPa P = 100 kPa P + P + P P = 85 kPa P = 8. A balloon contains 3 gases (hydrogen, oxygen, and carbon dioxide). If the balloon occupies a volume of 1.0 L , hydrogen gas has a pressure of 100 kPa, oxygen has a pressure of 85 kPa and carbon dioxide has a pressure of 115 kPa. What is the total pressure of the gases exerted on the balloon? PT = P = 100 kPa P + P + P H2 H2 O2 CO2 P = 85 kPa PT = O2 100 kPa + 85 kPa + 115 kPa P = 115 kPa CO2 PT = ____ kPa PT = 300 kPa

C H P V = n R T = = = P = 5,920 kPa = 5.92 x 103 kPa P = ____ V = P Given a 32.0 g sample of methane gas (CH4) , determine the pressure that would be exerted on a container with a volume of 850 cm3 at a temperature of 30 C. Note: Carbon = 12.0 g/mol and Hydrogen = 1.0 g/mol Use Ideal Gas Law when given grams. P V = n R T P = ____ 1000 mL = 1 L so… V = 12.0 x 1 CH4 1.0 x 4 + 4.0 = 16.0 g 850 cm3 = 850 mL = 0.850 P (0.850) = (2) (8.31) (303) mol CH4 n = 1 32.0 g CH4 = 2 mol x Solve for P 16 g CH4 R = 8.31 kPa•L mol•K kPa L P = = (0.850) (2) (8.31) (303) 6 C Carbon 12.01 1 H Hydrogen 1.01 (0.850) (0.850) T = 30°C + 273 = 303 K P = 5,920 kPa = 5.92 x 103 kPa Answer w/ 3 Sig. Figs. Calculator 2 x 8.31 x 303 ÷ 0.850 = 5,924.54 kPa

N P V = n R T = = = V = 60.0 L P = 95.0 kPa V (2) (8.31) (343) V = n = 10. Given the following balanced equation. N2(g) + 3 H2(g)  2 NH3(g) What volume of ammonia gas (NH3) would be produced at 95.0 kPa and 70 C if you were given a 28.0 g sample of nitrogen gas (N2)? 1 2 Use Ideal Gas Law when given grams. P V = n R T P = 95.0 kPa (95.0) = V (2) (8.31) (343) V = ______ 1 mol N2 mol NH3 n = 28.0 g N2 x x = 2 mol NH3 mol N2 Solve for V 28.0 g N2 R = 8.31 kPa•L mol•K L 7 N Nitrogen 14.01 N2 14.01 x 2 28.02g = = (95.0) V (2) (8.31) (343) (95.0) (95.0) T = 70°C + 273 = 343 K V = 60.0 L Answer w/ 3 Sig. Figs. Calculator 2 x 8.31 x 343 ÷ 95.0 = 60.0069 L