1.
A sample of CO
2
with a pressure of 76.0 mm Hg in a volume of 0.200 L is
compressed so that the new pressure of the gas is 90.0 mm Hg. What is the new
volume of the gas? Assume that temperature is constant.
a.
P
1
= 76.0 mm Hg; V
1
= 0.200 L ; P
2
= 90.0 mm Hg ; V
2
= ???
b.
P
1
V
1
= P
2
V
2
c.
V
2
= P
1
V
1
/ P
2
d.
V
2
= [(76 mm Hg)(.200 L)] / (90 mm Hg) = .
169 L
2.
Suppose you have a sample of CO
2
in a gastight syringe. The gas volume is 38.0
mL at 25˚C. What is the final volume of the gas if you hold the syringe in your
hand to raise its temperature to 40˚C?
a.
V
1
= 38 mL ; T
1
= 25˚C ; T
2
= 40˚C ; V
2
= ????
b.
T
1
= 25 + 273 = 298 K
c.
T
2
= 40 + 273 = 313 K
d.
V
1
/ T
1
= V
2
/ T
2
e.
V
2
= T
2
x (V
1
/ T
1
)
f.
V
2
= (313 K) x (38 mL / 298 K) =
40 mL
3.
You have a 34.0 L cylinder of helium at a pressure of 110 atm and a temperature
of 43˚C. How many balloons can you fill, each with a volume of 26.0 L, on a day
when the atmospheric pressure is 600 mm Hg and the temperature is 15˚C?
a.
V
1
= 34 L ; P
1
= 110 atm ; T
1
= 43˚C ; P
2
= 600 mm Hg ; T
2
= 15˚C ; V
2
= ???
b.
P
2
= 600 mm Hg x 1 atm / 760 mm Hg = .7895 atm
c.
T
1
= 43 + 273 = 316 K
d.
T
2
= 15 + 273 = 288 K
e.
General Gas Law
P
1
V
1
/ T
1
= P
2
V
2
/ T
2
f.
V
2
= (P
1
V
1
T
2
) / (T
1
P
2
)
g.
V
2
= (110 atm x 34 L x 288 K) / (316 K x .7895 atm) = 4317.57 L
h.
Volume of a balloon is 26L
i.
4317.57 / 26 = 166.06
j.
The number of balloons that can be filled are 166.
4.
A hot air balloon is filled with 2000 grams of H
2
. If the temperature of the gas is
32˚C and its pressure is 800 mm Hg, what is the volume of the balloon?
a.
2000 grams H
2
x 1 mol / 1.0079 g = 1984.3238 mol H
2
b.
32 + 273 = 305 K
c.
800 mm Hg x 1 atm / 760 mm Hg = 1.053 atm
d.
PV = nRT
e.
V = nRT / P
f.
V = [(1984.3238 mol H
2
)(.08206 (L∙atm)/(mol∙K))(305 K)] / (1.053 atm)
g.
V = 47181 L ≈47200 L
5.
A .306 gram sample of a gaseous compound has a pressure of 400 mm Hg in a
volume of 225 mL at 27˚C. What is its molar mass?
a.
Mass = .306 gram ; P = 400 mm Hg ; V = 225 mL ; T = 27˚C
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Density of the gas = grams /
Liter
i.
V = 225 mL x 1 L / 1000 mL = .225 L
ii.
d = .306 grams / .225 L = 1.36 g / L
c.
M
= dRT / P
i.
400 mm Hg x 1 atm / 760 mm Hg = .52634 atm
ii.
[(1.36 g / L)(.08206 (L∙atm)/(mol∙K))(300 K)] / (.52632 atm)
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 Spring '08
 LaDuca
 Mole, mol, mm Hg

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