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Chemistry 141 Lecture
Dr. Ben Tovrog
Chapter 5 Supplemental Problems  Answers
1.
An ideal gas has a volume of 18.25 L at a temperature of 15.9
°
C.
The temperature of the gas is
raised to 40.7
°
C while the number of moles and the pressure is kept constant.
What is the new volume
of the gas (in L)?
V
1 =
18.25 L
T
1
= 15.9 + 273.15 = 299.05 K
=
T
2
= 40.7 + 273.15 = 313.85 K
V
2
= ?
(18.25 L)/(299.05 K) = V
2
/ (313.85 K)
V
2
= 19.15 L.
2.
An ideal gas has a volume of 10.8 L at a temperature of 25.0
°
C and a pressure of 1.6 atm.
The pressure of the gas is reduced to 370 mmHg, but the temperature and and number of
moles of the gas are kept constant.
What is the new volume of the gas (in L)?
P
1
= 1.6 atm
P
2
= 370 mmHg/760 mmHg/atm = 0.487 atm
V
1 =
10.8 L
V
2 =
?
V
2
= P
1
V
1
/P
2
= (1.6 atm)*(10.8 L) / 0.487 atm = 35.48 L.
3.
Consider a 1.53 L sample of gaseous SO
2
at a pressure of 2.7 atm.
If the volume of the sample was
reduced to 0.57 L, what would the pressure of the sample be?
P
1
= 2.7 atm
P
2
= ?
P
2
= P
1
V
1
/V
2
= (2.7 atm)*(1.53 L)/ (0.57 L) =
7.25 atm
V
1 =
1.53 L
V
2 =
0.57 L
4. Draw a line from “Charles’ Law” and from “Boyle’s Law” to the correct description of each
gas law:
direct relationship between
V
and
P
Charles’ Law
direct relationship between
V
and
T
inverse relationship between
V
and
P
Boyle’s Law
inverse relationship between
V
and
T
5.
Fifteen grams of dry ice (CO
2
) are put into a balloon and allowed to sublime (turn to gas) at O C and
1 atm.
What will be the volume of the balloon when all the CO
2
has sublimed?
1 mole = 22.4 L.
15 g CO
2
/ 44.0 g/mole = 0.341 mol CO
2
V
1
V
2
____
T
1
T
2
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V = (0.341 mol) * (22.41 L/mol) = 7.64 L CO
2
gas.
6.
A sample of diborane gas (B
2
H
6
) has a pressure of 345 torr at a temperature of 15 C and a volume of
3.48 L.
If conditions are changed so that the temperature is 36 C and the pressure is 468 torr, what will
be the volume of the sample?
T
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 Summer '08
 Mulfod
 Mole

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