Chem 2 AP Homework #5-2: Problems pg. 204-205 #5.27, 5.29, 5.32, 5.34, 5.38-5.44 (even), 5.48, 5.50,
5.130*
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An ideal gas has P-V-T behavior that is completely accounted for by the ideal gas equation, PV=nRT.
Its molecules/atoms do not attract or repel each other (collisions are ideally elastic) and their volume is
negligible compared to the volume of their container.
29
STP is 0 °C (273.15 K) and 1 atm (101.3 kPa). At STP, the molar volume of an ideal gas is 22.41 L.
32
6.9 moles CO in 30.4 L at a temperature of 62°C
Because no changes in gas properties occur, we can use the ideal gas equation to calculate the pressure.
L⋅atm
(6.9 mol) (0.0821 mol⋅K
)(62 + 273)K = 6.2 atm
nRT
, so P =
P =
V
30.4 L
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In this problem, the moles of gas and the volume the gas occupies are constant (V1 = V2 and n1 = n2).
Initial conditions
T1 = (25 + 273)K = 298 K
P1 = 0.800 atm
P1
P
= 2
T1
T2
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→ T2 =
T1 P2
(298 K)(2.00 atm)
, or T2 =
= 745 K = 472°C
P1
(0.800 atm)
In this problem, the moles of gas and the pressure on the gas are constant (n1 = n2 and P1 = P2).
Initial conditions
T1 = (20.1 + 273) K = 293.1 K
V1 = 0.78 L
V1 V2
=
T1
T2
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Final Conditions
T2 = ?
P2 = 2.00 atm
→ V2 =
Final Conditions
T2 = (36.5 + 273)K = 309.5 K
V2 = ?
V1T2
(0.78 L)(309.5 K)
, or V2 =
= 0.82 L
T1
(293.1 K)
In the problem, temperature and pressure are given. If we can determine the moles of CO2, we can
calculate the volume it occupies using the ideal gas equation.
? mol CO2 = 88.4 g CO2 ×
VCO2
1 mol CO2
= 2.01 mol CO2
44.01 g CO2
L⋅atm
(2.01 mol ) (0.0821 mol⋅K
)(273 K ) = 45.1 L
nRT
=
=
P
(1 atm )
Alternatively, we could use the fact that 1 mole of an ideal gas occupies a volume of 22.41 L at STP:
22.41 L
? L CO2 = 2.01 mol CO2 ×
= 45.0 L CO2
1 mol
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The molar mass of CO2 = 44.01 g/mol. Since PV = nRT, we write:
P =
nRT
V
P =
(0.050 g ×
1 mol
44.01 g
) (0.0821
4.6 L
L⋅atm
mol⋅K
)(30 + 273)K
= 6.1 × 10 −3 atm
2
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HOMEWORK #5-2 ANSWER KEY
At 741 torr and 44°C, 7.10 g of a gas occupy a volue or 5.40 L. What is the molar mass of the gas?
Start with PM = dRT → M =
d =
M
dRT
P
7.10 g
1 atm
= 1.31 g/L , T = 44° + 273° = 317 K, P = 741 torr ×
= 0.975 atm
5.40 L
760 torr
(1.31 ) (0.0821
=
g
L
L⋅atm
mol⋅K
)(317 K) = 35.0 g / mol
0.975 atm
Alternatively, we can solve for the molar mass by writing:
mass of compound
molar mass of compound =
moles of compound
n =
PV
RT
→
n =
(0.975 atm )(5.40 L )
= 0.202 mol , so
L⋅atm
(0.0821 mol⋅K
)(317 K )
molar mass of compound =
mass of compound
7.10 g
=
= 35.1 g / mol
moles of compound
0.202 mol
48
Calculate the density of hydrogen bromide (HBr) gas in grms per liter at 733 mm Hg and 46°C.
PM
Start with PM = dRT → d =
RT
M = 1.008 g/mol + 79.90 g/mol = 80.91 g/mol, T = 46° + 273° = 319 K,
1 atm
P = 733 mmHg ×
= 0.964 atm
760 mmHg
d =
(0.964 atm)
(
319 K
80.91 g
1 mol
)×
mol ⋅K
= 2.98 g / L
0.0821 L ⋅atm
Alternatively, we can solve for the density by writing: density =
mass
volume
If we have 1 mole of HBr, the mass is 80.91 g.
L⋅atm
(1 mol) (0.0821 mol⋅K
)(319 K ) = 27.2 L , so
nRT
→ V =
V =
P
0.964 atm
density =
mass
80.91 g
=
= 2.97 g / L
volume
27.2 L
HOMEWORK #5-2 ANSWER KEY
50
EF is SF4. At 20°C 0.100 g occupies 22.1 mL at a pressure of 1.02 atm. What is the molecular
formula?
We need to find molar mass, M: M =
T(K) = 20° + 273° = 293 K
(
M =
0.100 g
22.1 mL
) (0.0821
L⋅atm
mol⋅K
dRT
P
)(20 + 273 K) = 107 g/mol
1.02 atm
empirical mass = 32.07 g/mol + 4(19.00 g/mol) = 108.07 g/mol
molar mass
≈ 1. Therefore, the molecular formula is the same as the empirical formula, SF4.
empirical mass
130
One oxide of nitrogen has a density of 1.33 g/L at 764 mm Hg and 150°C. What is its formula?
M =
(
)
L⋅atm
1.33 Lg (0.0821 mol⋅K
)(150 + 273)K = 45.9 g/mol
dRT
=
atm
P
764 mmHg × 7601mmHg
(
Some nitrogen oxides and their molar masses are:
NO 30 g/mol
N2O 44 g/mol
)
NO2
46 g/mol
The nitrogen oxide is most likely NO2, although N2O cannot be completely ruled out.
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