Name _______________________________________ Period __________
Chemistry HP Unit 6 Test A – Gases (Version 2)
Directions: Fill in the circle or "bubble" that corresponds to the letter of the answer you have chosen as being the best
among the choices provided.
1. Convert 2.30 atm into mm Hg.
(a) 2300 mm Hg
(b) 1750 mm Hg
(c) 2.30 mm Hg
(d) 0.00300 mm Hg
? 𝑚𝑚 𝐻𝑔 = 2.30 𝑎𝑡𝑚 ×
760 𝑚𝑚 𝐻𝑔
= 1750 𝑚𝑚 𝐻𝑔
1 𝑎𝑡𝑚
2. What volume of O2, collected at 22.0 ᵒC and 728 mm Hg would be produced by the decomposition of 8.15 g
KClO3? 2KClO3(s) --> 2KCl(s) + 3O2(g)
(a) 1.12 L (b) 1.48 L (c) 1.68 L
(d) 2.23 L (e) 2.52 L
We first need to solve for mol O2, and then plug that into the Ideal Gas Law to calculate volume.
? 𝑚𝑜𝑙 𝑂2 = 8.15 𝑔 𝐾𝐶𝑙𝑂3 ×
1 𝑚𝑜𝑙 𝑔 𝐾𝐶𝑙𝑂3
3 𝑚𝑜𝑙 𝑔 𝑂2
×
= 0.0998 𝑚𝑜𝑙 𝑂2
122.55 𝑔 𝐾𝐶𝑙𝑂3 2 𝑚𝑜𝑙 𝑔 𝐾𝐶𝑙𝑂3
We also need to convert 728 mm Hg to atm:
? 𝑎𝑡𝑚 = 728 𝑚𝑚 𝐻𝑔 ×
1 𝑎𝑡𝑚
= 0.958 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
𝐿 ∙ 𝑎𝑡𝑚
𝑛𝑅𝑇 (0.0998 𝑚𝑜𝑙) (0.0821 𝐾 ∙ 𝑚𝑜𝑙 ) (22 + 273.15)
𝑉=
=
= 2.52 𝐿 𝑂2
𝑃
0.958 𝑎𝑡𝑚
3. The density of chlorine (Cl2) gas at 25 °C and 60. kPa is __________ g/L.
(a) 0.86 (b) 1.7 (c) 4.9 (d) 20 (e) 0.58
First convert kPa to atm:
? 𝑎𝑡𝑚 = 60. 𝑘𝑃𝑎 ×
1 𝑎𝑡𝑚
= 0.59 𝑎𝑡𝑚
101.325 𝑘𝑃𝑎
𝑔
(70.90
) (0.59 𝑎𝑡𝑚)
𝑀𝑀 × 𝑃
𝑔
𝑚𝑜𝑙
𝑑=
=
= 1.7
𝐿 ∙ 𝑎𝑡𝑚
𝑅𝑇
𝐿
(0.0821
) (25 ℃ + 273.15 𝐾)
𝐾 ∙ 𝑚𝑜𝑙
4. Which graph best represents the pressure-volume relationship for an ideal gas at constant temperature?
(d) Is the correct answer.
5. A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of
nitrogen, argon, and helium were 44.0, 486, and 218 torr, respectively. The partial pressure of neon in the vessel
was __________ torr.
(a) 521 (b) 42.4 (c) 19.4 (d) 760 (e) 239
𝑃𝑡𝑜𝑡𝑎𝑙 = 44.0 𝑡𝑜𝑟𝑟 + 486 𝑡𝑜𝑟𝑟 + 218 𝑡𝑜𝑟𝑟 + 𝑃𝑁𝑒 = 987 𝑡𝑜𝑟𝑟
𝑃𝑁𝑒 = 239 𝑡𝑜𝑟𝑟
6. At what temperature will 41.6 grams N2 exert a pressure of 815 mm Hg in a 20.0 L cylinder?
(a) 134 K (b) 177 K (c) 238 K
(d) 337 K (e) 400 K
Convert 41.6 g N2 to mole:
? 𝑚𝑜𝑙 𝑁2 = 41.6 𝑔 𝑁2 ×
1 𝑚𝑜𝑙 𝑁2
= 1.48 𝑚𝑜𝑙 𝑁2
28.02 𝑔
? 𝑎𝑡𝑚 = 815 𝑚𝑚 𝐻𝑔 ×
1 𝑎𝑡𝑚
= 1.07 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
Convert mm Hg to atm:
Then, plug it into the Ideal Gas Law and solve for T:
𝑇=
(1.07 𝑎𝑡𝑚)(20.0 𝐿)
𝑃𝑉
=
= 177 𝐾
𝑛𝑅 (1.48 𝑚𝑜𝑙) (0.0821 𝐿 ∙ 𝑎𝑡𝑚 )
𝐾 ∙ 𝑚𝑜𝑙
7. What is the molecular weight of a gas if 0.104 gram of the gas occupies 48.7 mL at STP?
(a) 5.06 g/mol (b) 28.2 g/mol (c) 34.5 g/mol
(d) 40.0 g/mol (e) 47.9 g/mol
We first find the number of moles of gas, and then divide the number of grams by that number of moles. Since
this gas is at STP:
? 𝑚𝑜𝑙 = 48.7 𝑚𝐿 ×
1𝐿
1 𝑚𝑜𝑙
×
= 0.00217 𝑚𝑜𝑙
1000 𝑚𝐿 22.414 𝐿
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡 =
𝑔
0.104 𝑔
𝑔
=
= 47.9
𝑚𝑜𝑙 0.00217 𝑚𝑜𝑙
𝑚𝑜𝑙
8. A gas at a pressure of 10.0 N/m2 exerts a force of 5.0 N on an area of _________ m2.
(a) 5.0 (b) 50. (c) 2.0 (d) 0.50
𝐴=
𝐹
5.0 𝑁
=
= 0.50 𝑚2
𝑃 10.0 𝑁
𝑚2
9. What volume of CH4 at 0 ᵒC and 1.00 atm contains the same number of moles as 0.50 L of N2 measured at 27 ᵒC
and 1.50 atm?
(a) 0.37 L (b) 0.46 L (c) 0.68 L
(d) 0.50 L (e) 0.82 L
First, determine how many moles are in 0.50 L of N2 there are:
𝑛=
(1.50 𝑎𝑡𝑚)(0.50 𝐿)
𝑃𝑉
=
= 0.030 𝑚𝑜𝑙
𝑅𝑇 (0.0821 𝐿 ∙ 𝑎𝑡𝑚 ) (27 + 273.15)
𝐾 ∙ 𝑚𝑜𝑙
We know that 1 mole of a gas at STP occupies 22.414 L, so we figure out how many L 0.0304 mol of CH4 would
occupy:
? 𝐿 𝐶𝐻4 = 0.030 𝑚𝑜𝑙 ×
22.414 𝐿
= 0.68 𝐿
1 𝑚𝑜𝑙
10. A gas at a pressure of 15.0 Pa exerts a force of __________ Pa·m2 on an area of 5.0 m2.
(a) 75 (b) 3.0 (c) 7.5 (d) 55 (e) 0.33
𝐹 = 𝑃 × 𝐴 = (15.0 𝑃𝑎)(5.0 𝑚2 ) = 75 𝑃𝑎 ∙ 𝑚2
11. What are the standard temperature and pressure conditions for gases?
(a) 0 ᵒC and 0 mm Hg
(b) 0 K and 760 mm Hg
(c) -273 ᵒC and 1 atm
(d) 0 ᵒC and 760 mm Hg
(e) 0 ᵒC and 1 mm Hg
12. If 3.0 L of helium at 20.0 ᵒC is allowed to expand to 4.4 L, with the pressure remaining the same, what is the new
temperature?
(a) 702 ᵒC (b) 430 ᵒC (c) 157 ᵒC
(d) -30.0 ᵒC (e) -55 ᵒC
𝑉1 𝑉2
=
𝑇1 𝑇2
𝑇1 = 20.0 ℃ + 273.15 = 293.15 𝐾
𝑇2 =
𝑉2 𝑇1 (4.4 𝐿)(293.15 𝐾)
=
= 430 𝐾
𝑉1
3.0 𝐿
℃ = 430 𝐾 − 273.15 = 156.85 ℃
13. A gas vessel is attached to an open-end manometer filled with mercury as shown below.
The difference in heights of the liquid in the two sides of the manometer is 32.3 cm when the atmospheric
pressure is 755 mm Hg. Calculate the pressure of the enclosed gas.
(a) 1.42 atm
(b) 0.993 atm
(c) 432 mm Hg
(d) 0.425 atm
Since the arm that is attached to the gas is lower than the other, the gas is exerting a greater pressure than the
atmosphere, the difference between the two being the difference in height between the two arms.
𝑃𝑔𝑎𝑠 = 755 𝑚𝑚 𝐻𝑔 + 323 𝑚𝑚 𝐻𝑔 = 1078 𝑚𝑚 𝐻𝑔
? 𝑎𝑡𝑚 = 1078 𝑚𝑚 𝐻𝑔 ×
1 𝑎𝑡𝑚
= 1.418 𝑎𝑡𝑚
760 𝑚𝑚 𝐻𝑔
14. Which of the following contains the most molecules?
(a) 1.00 L of CH4 at 0 ᵒC and 1.00 atm
(b) 1.20 L of N2 at 0 ᵒC and 1.00 atm
(c) 1.00 L of O2 at 0 ᵒC and 1.00 atm
(d) 0.90 L of CO2 at 0 ᵒC and 1.00 atm
(e) 0.50 L of CO and 0 ᵒC and 1.00 atm
Since equal volumes of gases at the same conditions of temperature and pressure contain equal numbers of
moles, it follows that greater volume would contain greater numbers of moles. And, the greater number of
moles, the greater number of molecules. The greatest volume is 1.20 L of N2 at the same STP conditions as all
the other gases listed.
15. In a gas mixture of He, Ne, and Ar with a total pressure of 8.40 atm, the partial pressure of Ar is __________ atm
if the partial pressures of He and Ne are 1.50 and 2.00 atm, respectively.
(a) 4.90 (b) 0.417 (c) 3.50
(d) 6.90 (e) 11.90
𝑃𝑡𝑜𝑡𝑎𝑙 = 1.50 𝑎𝑡𝑚 + 2.00 𝑎𝑡𝑚 + 𝑃𝐴𝑟 = 8.40 𝑎𝑡𝑚
𝑃𝐴𝑟 = 4.90 𝑎𝑡𝑚
16. A quantity of gas has a volume of 250.0 liters at 17.0 ᵒC and 3.00 atm pressure. To what volume must the gas be
increased for the gas to be under STP conditions?
(a) 78.4 L (b) 88.5 L (c) 706 L
(d) 771 L (e) 797 L
STP = 1 atm (P2), 273.15 K (T2)
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
𝑉2 =
𝑃1 𝑉1 𝑇2 (3.00 𝑎𝑡𝑚)(250.0 𝐿)(273.15 𝐾)
=
= 706 𝐿
(17.0 + 273.15 𝐾)(1 𝑎𝑡𝑚)
𝑇1 𝑃2
17. Ammonia gas is synthesized according to the balanced equation: N2(g) + 3H2(g) --> 2NH3(g)
If 15.0 L of nitrogen gas are reacted with an excess of hydrogen at STP, how many liters of ammonia could be
produced?
(a) 5.00 L (b) 10.0 L (c) 15.0 L
(d) 20.0 L (e) 30.0 L
Volumes of gases are directly related at the same conditions of temperature and pressure, since they are
independent of the type of gas.
? 𝐿 𝑁𝐻3 = 15.0 𝐿 𝑁2 ×
2 𝑚𝑜𝑙 𝑁𝐻3
= 30.0 𝐿 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2
18. If the volume of a confined gas were doubled while the temperature remains constant, what change (if any)
would be observed in the pressure?
(a) It would be half as large.
(b) It would double.
(c) It would be four times as large.
(d) It would be ¼ as large.
(e) It would remain the same.
Pressure and volume are inversely related, which means that, if you double one, you halve the other.
19. What volume will 12.4 grams of CO2 occupy at STP?
(a) 6.32 L (b) 8.46 L (c) 4.42 L
(d) 11.7 L (e) 9.68 L
At STP, 1 mole of gas = 22.414 L:
? 𝐿 = 12.4 𝑔 𝐶𝑂2 ×
1 𝑚𝑜𝑙 𝐶𝑂2
22.414 𝐿
×
= 6.32 𝐿
44.01 𝑔
1 𝑚𝑜𝑙 𝐶𝑂2
20. Which of the following is equivalent to 1025 mm Hg?
(a) 1.249 atm
(b) 1.367 Pa
(c) 0.7415 atm
(d) 136.7 kPa
? 𝑘𝑃𝑎 = 1025 𝑚𝑚 𝐻𝑔 ×
101.325 𝑘𝑃𝑎
= 136.7 𝑘𝑃𝑎
760 𝑚𝑚 𝐻𝑔