Page 198 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 12] GASES 187

temperature, the resulting straight line goes through the origin, and thus volume and absolute temperature
are directly proportional.

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12.16. If 42.3 mL of gas at 22 C is changed to 44 C at constant pressure, what is its final volume?
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Ans. Absolute temperatures must be used:
22 C + 273 = 295 K
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◦
44 C + 273 = 317 K
◦
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According to Charles’ law:
V 1 V 2
=
T 1 T 2
(42.3mL)(317 K)
V 1 T 2
V 2 = = = 45.5mL
T 1 295 K
Note that the volume is not doubled by doubling the Celsius temperature.
12.17. If 0.979 L of gas at 0 C is changed to 737 mL at constant pressure, what is its final temperature?
◦
Ans. Using the reciprocal of the equation usually used for Charles’ law:
T 2 T 1
=
V 2 V 1
and solving for T 2 :
(273 K)(737 mL)
T 1 V 2
T 2 = = = 206 K
V 1 979 mL
The temperature was lowered to reduce the volume.

THE COMBINED GAS LAW
12.18. Calculate the missing value for each set of data in the following table:

P 1 V 1 T 1 P 2 V 2 T 2

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◦
(a) — 29.1 L 45 C 780 torr 2.22 L 77 C
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(b) 12.0 atm — 28 C 12.0 atm 750 mL 53 C
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(c) 721 torr 200 mL — 1.21 atm 0.850 L 100 C
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(d) 1.00 atm 4.00 L 273 K 1.00 atm 2.00 L —
(e) 7.00 atm — 333 K 3.10 atm 6.00 L 444 K
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(f ) 1.00 atm 3.65 L 130 C — 5.43 L 130 C
◦
Ans. Each problem is solved by rearranging the equation
P 1 V 1 P 2 V 2
=
T 1 T 2
All temperatures must be in kelvins.
(780 torr)(2.22 L)(318 K)
P 2 V 2 T 1
(a) P 1 = = = 54.1 torr
T 2 V 1 (350 K)(29.1L)
P 2 V 2 T 1 (12.0 atm)(750 mL)(301 K)
(b) V 1 = = = 692 mL
P 1 T 2 (12.0 atm)(326 K)
Since P did not change, Charles’ law could have been used in the form
V 2 T 1
V 1 =
T 2
P 1 V 1 T 2 (721 torr)(0.200 L)(373 K)
(c) T 1 = = = 68.8K
P 2 V 2 (1.21 atm)(760 torr/atm)(0.850 L)