# Review - Gases

Chem I | Chem II AP
1)
If 300 mL of nitrogen at 280 mmHg were suddenly compressed into a 140 mL volume, what would be the final pressure?
280 mmHg x
 300 mL 140 mL
= 600 mmHg

Volume decreases so pressure increases.

2)
The temperature of 546 K is equivalent to:
 546 K - 273 = 273°C
3)
The volume of a given sample of gas is 150.0 mL at 150°C and 1.0 atm pressure. Calculate the volume of the same gas sample at 10°C and 760 torr.
150 mL x
 283 K 423 K
= 100 mL

Temperature decreases so volume also decreases.

4)
A sample of Ar has a volume of 3.0 liters at 27°C. At what temperature would this sample have a volume of 5.0 liters? (P constant)
300 K x
 5.0 L 3.0 L
= 500 K
 500 K - 273 = 227 °C

Volume increases so temperature increases.

5)
If the pressure in a steel gas cylinder is 37.5 atm at 22.0°C, what would the pressure be at a temperature of 300.0°C, as might be created in a fire?
37.5 atm x
 573 K 295 K
= 72.8 atm

Temperature increases so pressure increases.

6)
If the gas cylinder in question 5 were to burst at a temperature of 140.0 atm or greater, above what temperature would the integrity of the gas cylinder be in question?
295 K x
 140.0 atm 37.5 atm
= 1101 K
 1101 K - 273 = 828 °C

This equation will find out the temperature when the pressure is 140 atm. Any temperatures greater than this will cause the container to burst.

7)
If 16.0g of He occupy 23.6 liters at some T and P, how many liters would 40.0 g of He occupy at the same T and P?
 16 g 23.6 L
=
 40 g x
x = 59 L

Because this is Helium at the same temperature and pressure, the grams and liters can be set up as a direct proportion.

Alternately, you could solve for moles and use Avogadro's Law; moles increases so volume increases.

8) Which one of the following statements is TRUE? The standard conditions of T and P are 25°C and 1.0 atm 70.9 g of chlorine gas at 0°C and 1.0 atm will occupy 22.4 L The volume of a mole of any gas is 22.4 L. STP conditions are 0 K and 1.0 atm. Choice A is not correct because standard temperature is 0°C.Choice C is not correct because the volume is 22.4 L.Choice D is not correct because standard temperature is 273 K.
9)
The density of a gas at STP was measured as 1.43 g/L. Which one of the following is a possible formula for the gas?
1.43 g/L =
 x 22.4 L
x = 32 g

Since the gas is at STP, 22.4 L will be the volume of a mole.

32 g is the g-molar mass of O2.

10)
If 1.12 L of an unknown gas has a mass of 3.55 g at STP, the g-MM of the gas is:
 3.55 g 1.12 L
=
 x 22.4 L
x = 71 g

Because this gas is at STP, 22.4 L is the volume of a mole.

11)
The volume of a gas weighing 0.55 g was found to be 210.0 mL when measured at 25°C and 740 torr. What is the g-MM of this gas?
MW =
 gRT PV
=
 (0.55 g)(0.0821 Latm/Kmol)(298 K) (0.974 atm)(0.21 L)
= 65.8 g/mol

Alternately, you could solve for n using PV=nRT then plug the moles into MW=mass/mol

12)
The density of N2 at STP is:
 28 g 22.4 L
= 1.25 g/L

Since this gas is at STP, the volume of one mole is at 22.4 L.

13)
The density of SO2 at 20.0°C and 740 torr is:
D =
 PMW RT
=
 (0.974 atm)(64.1 g/mol) (0.0821 Latm/Kmol)(293 K)
= 2.6 g/L

You cannot use 22.4 for the volume because the gas is not at STP.

14)
Calculate the volume of 2.00 moles of an ideal gas at a pressure of 0.100 atm and a temperature of 300.0 K.
V =
 nRT P
=
 (2 mol)(0.0821 Latm/Kmol)(300 K) 0.100 atm
= 492.6 L
15)
How many molecules are in 2.0 L of hydrogen gas at 17°C and 780 torr?
n =
 PV RT
=
 (1.03 atm)(2.0 L) (0.0821 Latm/Kmol)(290 K)
= 0.086 mol
0.086 mol x
 6.02 x 1023 mlc 1 mol
= 5.2 x 1022 mlc
16)
Given: 4 NH3 + 5 O2 → 4 NO + 6 H2
How many liters of NO are produced by the reaction of 10.0 liters of NH3 with 11.0 liters of O2? ( All gases at the same T and P)
10 L NH3 x
 4 mol NO 4 mol NH3
= 10 L NO
11 L O2 x
 4 mol NO 5 mol O2
=  8.8 L NO

Since the gases are at the same temperature and pressure, their volumes will cancel. You can skip the liters to moles and the moles to liters steps.

17)
What volume of H2, measured at STP, can be prepared by the electrolysis (decomposition) of 54.0 g of water?
 2 H2O → 2 H2 + O2
54 g H2O x
 1 mol 18 g
x
 2 mol H2 2 mol H2O
x
 22.4 L 1 mol
= 67.2 L
18)
Ten (10.0) grams of Na are placed into a flask containing 1.0 liter of Cl2 measured at STP. What is the maximum amount of NaCl that can form?
 2 Na + Cl2 → 2 NaCl
10 g Na x
 1 mol 23 g
x
 2 mol NaCl 2 mol Na
x
 58.5 g 1 mol
= 25.4 g NaCl
1.0 L x
 1 mol 22.4 L
x
 2 mol NaCl 1 mol Cl2
x
 58.5 g 1 mol
=  5.2 L NaCl

Since the gas is at STP the volume of one mole is 22.4 L.

19)
In a gaseous mixture of 1.0 mole of N2 and 4.0 moles of O2, the total pressure is 0.20 atm. The partial pressure of N2 is.
 1.0 mol 5.0 mol
=
 x 0.20 atm
x = 0.04 atm

Moles are directly proportional to the pressure.

20)
The rate of diffusion (effusion) of SO2 relative to the rate of diffusion of O2 at the same T and P is:
 VSO2 VO2
=
 32 64
= 0.71

SO2 diffuses 0.71 times as fast as O2.

21)
If a gas diffuses twice as fast as SO2 under the same conditions, its g-MM is:
 Vunknown gas VSO2
=
 64 x
= 2 x = 16
22) Which one of the following statements regarding the kinetic molecular theory of gases is incorrect? Gas molecules are in constant motion. Gas molecules are widely separated from one another. Gas molecules have zero volume. All gas molecules have the same velocity at the same temperature. When at the same temperature gases have the same kinetic energy, but they do not have the same velocity which is based on their molecular weight.
23) Gases behave ideally at All T and P High T and low P Low T and high P High T and high P
24)
Given: N2 + 3 H2 → 2 NH3
How many liters of NH3 will be produced under 2.0 atm and 30°C, when 56.0 g of nitrogen react with excess hydrogen?
 N2 + 3 H2 → 2 NH3
56 g N2 x
 1 mol 28 g
x
 2 mol NH3 1 mol N2
x
 12.44 L 1 mol
=  49.7 L NH3
V =
 nRT P
=
 (1 mol)(0.0821 Latm/Kmol)(303 K) 2.0 atm
= 12.44 L

Since the gas is not at STP we cannot use 22.4 L for the volume of a mole. The volume must be calculated using PV=nRT.

25)
Given: N2 + 3 H2 → 2 NH3
How many grams of nitrogen are required to produce 5.0 liters of NH3 under the conditions of 3.0 atm and 100°C?
 N2 + 3 H2 → 2 NH3
5.0 L NH3 x
 1 mol 10.21 L
x
 1 mol N2 2 mol NH3
x
 28 g 1 mol
=  6.86 g N2
V =
 nRT P
=
 (1 mol)(0.0821 Latm/Kmol)(373 K) 3.0 atm
= 10.21 L

Since the gas is not at STP we cannot use 22.4 L for the volume of a mole. The volume must be calculated using PV=nRT.

26) Given 4 g He in a 10 L container and 44 g CO2 in a 10 L container Which statement is correct if both containers are at STP? The gas in each container has the same density The velocity of the gas in each container is the same The average K.E. of each gas is the same. There are ten times as many He atoms as CO2 molecules. Choice A is not correct because in the same volume CO2 weighs 11 times more than He.Choice B is not correct because Helium has a lower molecular weight and therefore has a higher velocity.Choice D is not correct because there are the same number of molecules in each container because they are at the same temperature, pressure and volume.
27)
Given: 2 moles of Ne and 3 moles of He in a 8 L container at 10.0 atm. Which statement is correct?

Choice B is not correct because the gases have a pressure based on the number of moles in the container.

Choice C is not correct because Ne has a pressure of 4 atm.

Choice D is not correct because He has a pressure of 6 atm.

 2 mol Ne 5 mol total
=
 x 10 atm total
x = 4 atm Ne
 3 mol He 5 mol total
=
 x 10 atm total
x = 6 atm He