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Chapter 10 Exercise Key
1
Chapter 10 Exercise Key
Exercise 10.1 – Using the Ideal Gas Equation:
Krypton gas does a better job than
argon of slowing the evaporation of the tungsten filament in an incandescent light bulb.
Because of its higher cost, however, krypton is used only when longer life is considered
to be worth the extra expense.
a.
How many moles of krypton gas must be added to a 175mL incandescent light
bulb to yield a gas pressure of 117 kPa at 21.6
°
C?
( )
3
117 kPa 175 mL
PV
1 L
PV = nRT
n =
=
8.3145 L
kPa
RT
10 mL
294.8 K
Km
o
l
⎛⎞
⎜⎟
•
⎝⎠
•
=
8.35 × 10
−
3
mol Kr
b.
What is the volume of an incandescent light bulb that contains 1.196 g of Kr at a
pressure of 1.70 atm and a temperature of 97
°
C?
V
=
?
g
=
1.196 g
T
=
97
°
C + 273.15
=
370 K
P
=
1.70 atm
g
PV =
RT
M
0.082058 L • atm
1.196 g Kr
370 K
gRT
K•mole
V =
=
g
PM
1.70 atm 83.80
mole
=
0.255 L Kr
c.
What is the density of krypton gas at 18.2
°
C and 762 mmHg?
g
V
= ?
P = 762 mmHg
T = 18.2
°
C + 273.15 = 291.4 K
g
PV =
RT
M
()
g
762 mmHg 83.80
g
PM
1 atm
mole
=
=
L atm
V
RT
760 mmHg
0.082058
291.4 K
K mole
⎜
⎟
=
3.51 g/L
Copyright 2004 Mark Bishop
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View Full DocumentChapter 10 Exercise Key
2
Exercise 10.2 – Using the Combined Gas Law Equation:
A helium weather
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 Fall '06
 Mark

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