CHEMISTRY 161
HW CH6A
SU12
32, 34, 42, 44, 46, 50, 54, 58, 72, 74, 78, 82
6-32
On the square face:
P



38.6 g

1 kg
1000 g

9.8 m s
2


1.00 cm 2


1 m
100 cm

2
 
(b) On the rectangular face:
P



38.6 g

1 kg
1000 g
2.00 cm 2


1 m
100 cm

2
6-34
2


 
(a) 0.541 atm

760 mmHg

1 atm
411 mmHg
(b) 2.8 kPa

1 atm

760 mmHg

21 mmHg
101.325 kPa 1 atm
6-42
Let’s assume the initial pressure is 1 atm and the original volume is 1 L. The volume after compression ( V
2
) will be 0.10 L. Using the equation to solve for P
2
gives
PV

P V
2
1 atm 1 L

P
2

0.10 L
P
2

10 atm
The pressure increases by a factor of 10.
6-46
1.0 atm

6 L

12.5 atm
V
2

0.5 L

V
2
6-50

The line for 1.00 mol He is shown in gray and the line for 0.50 mol He is shown in black. Yes, the prediction from Problem 6.49 matches. We see that the volume of the 0.50 mol sample of He at the same T and P is exactly one-half that of the 1.00 mol He sample.
6-54
To find T
2
, use Charles’s law for this problem:
V
1

V
2
T
1
T
2 where V
1
= 5.6 L, T
1
= 78˚C (351 K), and V
2
= 4.3 L.
6-58
(a) If we double the volume by doubling the amount of gas in the sample, the percent volume increase is 100%.
(b) The percent change in the temperature is
1373 K – 517 K

100
 
166%
517 K
Therefore, the increase in temperature (b) produces the greatest increase in the volume for the gas sample.
6-72
Volume the vapor occupies:
Moles H O

100 g H O

1 mol
18.02 g

5.55 mol

179 L V

1.00 atm
6-74
623 mmHg

263 K
2.00 L

688 mmHg
268 K

1.85 L

V
2
V
2
6-78
(a) Moles of CO
2
produced in the reaction n

(b) Mass of CaCO
3
consumed in the reaction
 
298 K

2


2

1 mol CaCO
3 
100.1 g CaCO
1 mol CO
2
1 mol

0.0614 g CaCO
3
6-82 n
O
2

 
293 K

4.16 mol O
2
Mass of KO
2
required:
4.16 mol O
2

4 mol KO
2
3 mol O
2

71.10 g KO
2
1 mol

394 g KO
2