Gas Laws
Standard
Temperature and
Pressure (STP)
Standard temperature:
0°C or 273 K
Standard pressure:
1 atm, 760 mm Hg, or 101.3 kPa
Boyle’s Law
The pressure of a dry
gas is inversely
proportional to its volume
if the temperature is held
constant.
Inverse
Proportion
two terms related so that,
if one term increases,
the other term decreases
Boyle’s Law
P ,V
PV = k
P1V1 = P2V2
As pressure
increases,
volume
decreases.
Sample Problem 1
A sample of gas occupies 352
mL at a pressure of 3.17 atm.
If the P is reduced to 1.24 atm,
what volume will result?
P1V1 = P2V2
(3.17 atm)(352 mL) = (1.24 atm)V2
900 mL = V2
Question
A sample of gas occupies 200
mL at a P of 2.15 atm. If the P is
reduced to 1.25 atm, what is V2?
1.
2.
3.
4.
344 mL
537 mL
98 mL
89 mL
(2.15 atm)(200 mL) =
(1.25 atm) V2
430 mL = 1.25 V2
344 mL = V2
Question
A sample of gas is at a P of 700
mm Hg with a V of 300 mL.
What P will increase the V to
932 mL?
1.
2.
3.
4.
2,175 mm Hg
225 mm Hg
339 mm Hg
1 mm Hg
(700 mm Hg)(300 mL) =
P2(932 mL)
210,000 mm Hg = 932 P2
225 mm Hg = P2
Charles’s Law
The volume and absolute
temperature of a gas
are directly proportional to
one another if the
pressure is held constant.
Direct
Proportion
two terms related so that,
if one term increases, the
other term also increases
Charles’s Law
V ,T
V
=
k
T
V1
V2
=
T1
T2
As temperature
increases,
volume
increases.
Sample Problem 2
If a sample of gas occupies
478 mL at 126 K, what volume
will result if the gas is heated
to 245 K?
Sample Problem 2
V1
V2
=
T1
T2
(478 mL)
V2
=
(126 K)
(245 K)
(478 mL)(245) = (126)V2
929 mL = V2
Sample Problem 3
If a sample of gas occupies
478 mL at 126°C, what volume
will result if the gas is heated
to 245°C?
Sample Problem 3
V1 V2
=
T1 T2
(478 mL)
V2
=
(126°C)
(245°C)
NOTE: T must be in K—add 273
(478 mL)(518) = (399)V2
621 mL = V2
Question
If a sample of gas occupies
232 mL at 321 K, what V will
result if the gas it heated to
554 K?
1.
2.
3.
4.
134 mL
767 mL
400 mL
0.001 mL
232 mL
=
321 K
V2
554 K
(232 mL)(554 K) = (321 K)V2
128,528 mL = (321)V2
400 mL = V2
Question
If a sample of gas occupies 112
mL at 313°C, what will be the T
if the V becomes 52 mL?
REMEMBER KELVIN!
1.
2.
3.
4.
19 K
10 K
145 K
272 K
112 mL
=
586 K
52 mL
T2
(112 mL)T2 = (586 K)(52 mL)
112 T2 = 30,472 K
T2 = 272 K
Gay-Lussac’s Law
The pressure of a
confined gas is directly
proportional to its
absolute temperature
if the volume is held
constant.
Gay-Lussac’s Law
P ,T
P
=
k
T
P1
P2
=
T1
T2
As temperature
increases,
pressure
increases at a
constant
volume.
Question
If a the P of a tire is 35 psi at
21°C, what will the P be if the T
is raised to 41°C?
1.
2.
3.
4.
18 psi
33 psi
37 psi
68 psi
35 psi
294 K
=
P2
314 K
(35 psi)(314 K) = (294 K)P2
10,990 psi = 294 P2
37.4 psi = P2
Combined Gas Law
Combines Boyle’s law,
Charles’s law,
and Gay-Lussac’s law
Combined Gas Law
Boyle’s
PV
V
Charles’s
T
P
Gay-Lussac’s T
Combined Gas Law
P1V1
P2V2
=
T1
T2
Sample Problem 4
If a gas occupies 1050 L at a P
of 720 mm Hg and a T of 250 K,
what V will it occupy at a P of
600 mm Hg and a T of 800 K?
P1V1
P2V2
=
T1
T2
Sample Problem 4
If a gas occupies 1050 L at a P
of 720 mm Hg and a T of 250 K,
what V will it occupy at a P of
600 mm Hg and a T of 800 K?
720
(mm Hg)(1050 L)
=
600
(mm Hg)(V2)
(250 K)
(800 K)
(720)(1050)(800) = (250)(600)V2
Sample Problem 4
If a gas occupies 1050 L at a P
of 720 mm Hg and a T of 250 K,
what V will it occupy at a P of
600 mm Hg and a T of 800 K?
V2 = 4,032 L
Sample Problem 5
If a gas with V of 45 L at 700
mm Hg and 300 K changes
to STP, what is its new V?
700
760
(mm Hg) (45 L) (mm Hg)(V2)
=
(300 K)
(273 K)
(700)(45 L)(273) = (760)(300)V2
Sample Problem 5
If a gas with V of 45 L at 700
mm Hg and 300 K changes
to STP, what is its new V?
37.7 L = V2
Question
If a gas occupies 324 mL at a P
of 526 torr and a T of 322 K,
what V will it have at 433 torr
and 221 K?
1.
2.
3.
4.
270 mL
1,037 mL
389 mL
1,024 mL
(526 torr)(324 mL)
(322 K)
(433 torr)V2
=
(221 K)
(526)(324)(221) = (322)(433)V2
37,663,704 mL = 139,426 V2
270 mL = V2
Dalton’s Law of
Partial Pressures
The total pressure of a
mixture of gases equals
the sum of the partial
pressures.
Dalton’s Law of
Partial Pressures
PT = P1 + P2 + P3 + …
=
+
+
Vapor Pressure
the pressure exerted
by the vapor emitted
by a liquid or a solid
Vapor Pressure
Patm = Pgas + PH O
2
Patm − PH O = Pgas
2
Question
A gas is collected over water.
The P is 748 mm Hg. The T is
30°C. What is the partial P of
the gas?
1.
2.
3.
4.
32 mm Hg
780 mm Hg
716 mm Hg
Not enough information
Patm − PH O = Pgas
2
748 torr – 32 torr = 716 torr