Supplementary Problems
Multiple Choice and Short Answer:
1.
Answer: a factor of ½.
2.
Answer: 236 K; 423 K; 272.4 K
3.
Answer: 1.97 atm; 0.616 atm; 0.976 atm
4.
Answer: c; use Boyle’s Law to calculate V2.
5.
Answer: b; 1520 mm Hg is equivalent to 2.0 atm.
6.
Answer: e; convert 722 mm Hg to 0.950 atm; (153/0.95) × 13.2 L = 2.13 × 103 L.
7.
Answer: 182 K; 273 K; increase; 3.0 L
8.
Answer: c; convert to Kelvin; 328 K, 276 K; (276/328) × 13.3 L = 11.2 L.
9.
Answer: c; convert to atm; 1.20 atm (no change); (467/298) × 405 mL = 635 mL.
10. Answer: b; convert to Kelvin 293 K, 308 K; (308/293) × 1180 mm = 1240 mm.
11. Answer: d; convert to Kelvin 298 K; (950/275) × 298 K = 1029 K, or 756 oC.
12. Answer: c; convert to Kelvin 273 K; (3.0/1.05) × 273 K = 780 K, or 507 oC.
13. Answer: c; convert to Kelvin 273 K; (3.0/1.05) × 273 K = 780 K, or 507 oC.
14. Answer: b; convert to Kelvin 294 K, 225 K; (225/294)(745/63.1) × 1.41 × 104 L = 1.27 × 105 L.
15. Answer: d; convert to Kelvin 223 K, 323 K; (323/223)(2.00/4.00) × 700 mm = 507 mm.
16. Answer: d; conversions; 277 K, 293 K, 0.997 atm; (293/277)(6.4/.997) × 2.5 mL = 17 mL.
17. Answer: c; convert to Kelvin 253 K, 330 K; (330/253)(2.00/4.50) × 0.125 atm = 0.0725 atm.
18. Answer: d; conversions; 293 K, 312 K, 0.947 atm; (312/293)(.947/1) × 0.75 L = 0.76 L.
19. Answer: d; use PV = nRT to solve for T1 = 300 K; (0.25/0.5)(0.5/1) × 300 K = 75 K, or -198 oC.
20. Answer: e; use PV = nRT to solve for T; (200 × 2.24)/(10 × 0.0821) = 546 K, or 273 oC.
21. Answer: b; convert to Kelvin 318 K & 10.31 moles; (10.31 × 0.0821 × 318)/50 = 5.38 atm.
22. Answer: d; conversions; 293 K, 1.32 × 10-4 atm; (1.32 × 10-4 × 0.0010)/(293 × 0.0821) = 5.5 × 10-9 moles.
23. Answer: c; conversions; 296 K; (154 × 0.334)/(296 × 0.0821) = 2.12 moles.
24. Answer: b; conversions; 298 K, 3896 moles H2 or 2598 moles NH3; (2598 × 0.0821 × 298)/515 = 123 L.
25. Answer: b; conversions; 293 K, 0.98 atm, 0.267 moles H2O or 0.267 moles H2; (0.267 × 0.0821 × 293)/0.98 = 6.6 L.
26. Answer: c; one mole of silane gives two moles of hydrogen gas… under the same conditions, this will give twice the
volume.
27. Answer: d; convert to Kelvin 298 K; (1.00 × 1.00)/(298 × 0.0821) = 0.0409 moles oxygen, or 0.0894 moles peroxide…
3.04 g.
28. Answer: e; conversions; 273 K; (1.00 × 0.325)/(273 × 0.0821) = 0.0145 moles hydrogen, or 0.00966 moles Al… 3.04 g.
29. Answer: d; conversions; 293 K, 0.974 atm; (0.974 × 0.927)/(293 × 0.0821) = 0.0375 moles hydrogen/zinc. This is 2.45 g.
30. Answer: (1690 × 7.36)/1440 = 8.65 L
31. Answer: (6.09 × 4.51)/11.5 = 2.39
32. Answer: (6380 × 294.9)/8040 = 234.0
33. Answer: (7.400 × 0.432)/0.898 = 3.56
34. Answer: (0.700 × 2.42 × 223)/(1.15 × 1.15) = 286K, or 13.00 oC
35. Answer: (1.97 × 1.76 × 750)/(720 × 2.33) = 1180 mm Hg
36. Answer: (0.0443 × 0.0821 × 361)/2.33 = 0.563 atm
37. Answer: (1220 × 7.87)/(62.4 × 236) = 0.649 moles
38. Answer: grams = 100 liters × (2 g/22.4 liters) = 8.9 g.
39. Answer: grams = 22.4 liters × (3.24 g/1 liter) = 72.6 g.
40. Answer: liters = 2.6 moles × (22.4 liters/l mole) = 58.2 liters
41. Answer: moles = 100 liters × ( 1 mole/22.4 liters) = 4.5 m
42. Answer: liters = 80 g × (22.4 liters/32 g) = 56.0 liters
43. Answer: liters = 111 g × (22.4 liters/36.5 g) = 68.1 liters
44. Answer: molecules = 15.6 liters × (6 × 1023 molecules/22.4 liters) = 4.2 × 1023 molecules
45. Answer: O2 molecules = .01 × l liter × (6 × 1023 molecules/22.4 liters) = 2.68 × 1020 molecules
46. Answer: liters = 37.5 g × x (22.4 liters/28 g) = 30 liters
47. Answer: molecules = 57 g × (6 x 1023 molecules/44 g) = 7.8 × 1023 molecules
48. Answer: grams = 22.4 liters × (3.90 g/1 liter) = 87.4 g
49. Answer: 22.4 liters × (32 g/47 liters) = 15.3 g
50. Answer: grams = 1 molecule × (28 g/6 × 1023 molecules) = 4.7 × 10-23 g
In-Chapter Problems
Exercise 9.1 Pressure-Volume Relationships
a)
The pressure of 12.5 L of a gas is 0.82 atm. If the pressure changes to 1.32 atm, what will the final volume be?
P1V1 = P2V2 ;
V2 =
( 0.82 atm ) (12.5 L )
1.32 atm
= 7.8 L
b) A sample of helium gas has a pressure of 860.0 mm Hg. This gas is transferred to a different container having a volume
of 25.0 L; in this new container, the pressure is determined to be 770.0 mm Hg. What was the initial volume of the
gas?
P1V1 = P2V2 ;
V1 =
( 25.0 L ) ( 770.0 mm )
( 860.0 mm )
= 22.4 L
Exercise 9.2 Temperature-Volume Relationships
a)
A 50.0 mL sample of gas at 26.4˚ C, is heated at constant pressure until its volume is 62.4 mL . What is the final
temperature of the gas?
b) A sample container of carbon monoxide occupies a volume of 435 mL at a temperature of 298 K. What would its
temperature be if the pressure remained constant and the volume was changed to 265 mL? (182 K)
V1 V2
= ;
T1 T2
Exercise 9.3 Ideal Gas Law Calculations
T2 =
( 265 mL ) ( 298 K ) = 182 K
435 mL
a)
A 0.0500 L sample of a gas has a pressure of 745 mm Hg at 26.4˚ C. The temperature is now raised to 404.4 K and the
volume is allowed to expand until a final pressure of 1.06 atm is reached. What is the final volume of the gas?
b)
c)
P1V1 P2V2
=
;
n1T1 n2T2
V2 =
( 745 mm ) ( 0.0500 L ) ( 404.4 K ) = 0.0625 L
( 299.4 K ) ( 805.6 mm )
When 128.9 grams of cyclopropane (C3H6) are placed into an 8.00 L cylinder at 298 K, the pressure is observed to be
1.24 atm. A piston in the cylinder is now adjusted so that the volume is now 12.00 L and the pressure is 0.88 atm.
What is the final temperature of the gas?
P1V1 P2V2
=
;
n1T1 n2T2
T2 =
( 0.88 atm ) (12.00 L ) ( 298 K ) = 317 K
( 8.00 L ) (1.24 atm )
Exercise 9.4 Ideal Gas Law Calculations: Reaction Stoichiometry
a)
When Fe2O3 is heated in the presence of carbon, CO2 gas is produced, according to the equation shown below. A
sample of 96.9 grams of Fe2O3 is heated in the presence of excess carbon and the CO2 produced is collected and
measured at 1 atm and 453 K. What volume of CO2 will be observed?
2 Fe2O3(s) + 3 C (s)
96.9 g
= 0.607 moles;
159.69 g
mol
PV = nRT ; V =
4 Fe (s) + 3 CO2 (g)
( 0.607 moles ) ( 0.0821 L atm mol-1 K -1 ) ( 453 K )
1 atm
= 22.6 L
b) The reaction of zinc and hydrochloric acid generates hydrogen gas, according to the equation shown below. An
unknown quantity of zinc in a sample is observed to produce 7.50 L of hydrogen gas at a temperature of 404 K and a
pressure of 1.75 atm. How many moles of zinc were in the sample?
Zn (s) + 2 HCl (aq)
PV = nRT ; n =
( 0.0821 L atm mol
ZnCl2 (aq) + H2 (g)
-1
)
K -1 ( 404 K )
( 7.50 L ) (1.75 atm )
= 2.53 moles