Chemistry 121
Mines
Detailed Solutions to: Group Problems—Gases II
1
Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the
pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at
69.5°C.
Answer: 9.42 atm
Brief Work: P = nRT/V = 1.82 mol x 0.08206 L·atm·mol-1·K-1 x (69.5 + 273.15) K/5.43 L
2
What is the volume (in liters) occupied by 7.40 g of CO2 at STP?
Answer: 3.77 L
Brief Work: V = nRT/P = (7.40 g/44.0 g/mol) x 0.08206 L·atm·mol-1·K-1 x 273.15 K/1 atm
3
Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What
volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature
raised from 20oC to 35oC?
Answer: 7570 L
Brief Work: V2 = V1 x (P1/P2) x (T2/T1) [or P1V1/T1 = P2V2/T2]
= (49.0 L) x (150.0 atm/1.02 atm) x (308.15 K/293.15 K)
4
An inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, is
allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that
the temperature remains constant, what is the final volume of the balloon?
Answer: 1.4 L
Brief Work: V2 = V1 x (P1/P2) = 0.55 L x (1.0 atm/0.40 atm)
(problems below are from McMurry and Fay’s Chemistry, 4th Ed.)
5. 9.47. Which sample contains more molecules: 2.50 L of air at 50°C and 750 mm Hg
pressure or 2.16 L of CO2 at –10°C and 765 mm Hg pressure?
Answer: The 2.16 L sample.
Strategy:
1) Use the ideal gas law to find the number of moles of (molecules of) gas in each sample.
(NOTE: Make sure that T is in Kelvins and P is in atm (if you want to use the value of
0.08206 L⋅atm/(mol⋅K)).
2) Compare the two values--whichever one has more moles of molecules has the larger
number of molecules (because a mole of molecules is a fixed number [Avogadro's
number] of things).
Execution of Strategy:
1) (I'm assuming 3 sig figs in 750 mmHg and two sig figs in 50°C.)
⎛ 750 mmHg ⎞
⎜
⎟ ( 2.50 L )
760 mmHg/atm ⎠
PV
⎝
PV = nRT ⇒ n =
=
= 0.093079.. = 0.0931 mol
RT (0.08206 L ⋅ atm/(mol ⋅ K))(50. + 273 K)
⎛ 765 mmHg ⎞
⎜
⎟ ( 2.16 L )
760 mmHg/atm ⎠
PV
⎝
n=
=
= 0.1007.. = 0.101 mol
RT (0.08206 L ⋅ atm/(mol ⋅ K))(-10. + 273 K)
2) The second sample contains a greater number of moles and thus a greater # of
molecules.
6
9.27. Assume that you have a sample of gas at 350 K in a sealed container, as
represented in (a). Which of the drawings (b) – (d) represents the gas after the temperature
is lowered from 350 K to 150K?
(a)
(b)
(c)
(d)
Answer: (c)
Explanation:
ASSUMING THE GAS HAS NOT TURNED INTO A LIQUID (not stated, but implied), the
answer is (c) because no matter how low the temperature, as long as a gas is still a gas, the
particles move in straight lines until they hit something—that is, they will spread out to fill the
container in which they are enclosed. The effect of lowering the T at constant V (rather than
constant T, as in 9.26) is that the particles will move more slowly on average. This is not
apparent in a motionless picture as in this problem, but it should be imagined! (If the gas does
condense into a liquid at a T above 150 K, then either (b) or (d) could represent the sample,
although it looks like the space between the molecules in (b) is somewhat large for a liquid.)
7
9.29. Three bulbs, two of which contain different gases and one of which is empty, are
connected as shown in the following drawing:
Redraw the apparatus to represent the gases after the stopcocks are opened and the system
is allowed to come to equilibrium.
Answer:
Explanation: Gas particles move in one direction until they collide with something. That
means that if you wait long enough, all particles will spread out into the volume available, and
all particles will "mix" completely if given enough time. So, on average, the composition of
each "bulb" in the apparatus above will become equivalent after enough time has passed.
8
9.69. Ammonium nitrate can decompose explosively when heated according to the
equation:
2 NH4NO3(s) → 2 N2(g) + 4 H2O(g) + O2(g)
How many liters of gas would be formed at 450°C and 1.00 atm pressure by
explosion of 450 g of NH4NO3?
Answer: 1170 L (or 1.17 x 103 L)
Strategy: Recognize that this problem is just a stoichiometry problem that involves gases
(although it does have one little “twist”) You are asked about an amount of product
(“gases”) that would from when a given amount of reactant reacts. Convert the grams of
NH4NO3(s) into moles, use the mole ratios given by the balanced equation to figure out the
moles of gases that would be produced, and then use the ideal gas equation with the T and
P specified to calculate the volume that would be occupied that those gases. The “twist” or
tricky part of this problem is that they did not ask for a specific product gas, but rather (all)
“gas(es)”. So you could either use the mole ratios of each product and then add up all the
moles produced at the end, or you can figure out the ratio of “(total) moles of gas per mol of
NH4NO3 reacted”. I’ll do the latter in this key.
Execution: The molar mass of NH4NO3 is 80.04 g/mol. Thus 450 g is roughly 6 moles. To
get the precise value:
450 g
= 5.622 mol NH4NO3
80.04 g/mol
(where I’ve assumed that 450 g has three significant figures even
though strictly speaking the rule would indicate 2 here)
From the balanced equation, for every 2 moles of NH4NO3 that react, 2 moles of N2(g), 4
moles of H2O(g), and 1 mole of O2(g) are produced. That means that seven moles of
gas(es) are produced in total for every 2 moles of NH4NO3 that react. Thus if 5.622 mol of
NH4NO3 react, 7/2 of 5.622 moles of gas(es) will be formed:
⎛
⎞
7 moles gas form
⎟ = 19.677 moles of gas are made
⎝ 2 mol NH4NO3 that react ⎠
( 5.612 mol NH4NO3 ) ⎜
PV = nRT ⇒ V =
nRT (19.677 mol)(0.08206 L ⋅ atm/(mol ⋅ K))(450 + 273.16 K)
=
= 1167.68.. = 1170 L
P
(1 atm )