The Behavior of Gases
The Gas
Laws
STP
 Standard
Temperature and
Pressure:
 273 K and 760 mm Hg
 Or 0 C and 1atm
Temperature Conversions
= K − 273
 [°C] = 5/9(°F-32)
 [°C]
 [K]
= °C + 273
[K] = (°F + 459.67) × 5⁄9
= K × 9⁄5 − 459.67
 [°F] = °C × 9⁄5 + 32
 [°F]
A. Boyle’s Law
P
Volume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0
20.0
30.0
40.0
760.0
379.6
253.2
191.0
7.60 x 103
7.59 x 103
7.60 x 103
7.64 x 103
PV = k
V
A. Boyle’s Law
 The
pressure and volume
of a gas are inversely
related
• at constant mass & temp
P
PV = k
V
A. Boyle’s Law
Boyle’s Law Practice Problem
 Given
1.53 L of a sample of SO2 at a
pressure of 5.6 x 103 Pa. If the pressure
is changed to 1.5 x 104 Pa at a constant
temp, what will be the new volume of the
gas.
GIVEN:
WORK:
V1 = 1.53 L
P 1V 1= P 2V 2
P1 = 5.6 x 103 Pa 5.6 x 103 Pa x 1.53 L
V2 = ?
1.5 x 104 Pa
P2 = 1.5 x 104 Pa
= 0.57L
Gas Law Problems
A
gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1 = P2V2
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
B. Charles’ Law
V
T
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0
44.0
47.7
51.3
273.2
298.2
323.2
348.2
0.146
0.148
0.148
0.147
V
k
T
B. Charles’ Law
 The
volume and absolute
temperature (K) of a gas
are directly related
• at constant mass &
pressure
V
T
V
k
T
B. Charles’ Law
Charles’ Law Practice
 A sample
of gas at 15 degrees C and 1
atm has a volume of 2.58 L. What volume
will this gas occupy at 38 degree C and 1
atm? (Hint** convert to K first)
GIVEN:
V1 = 2.58 L
T1 = 15 C
V2 = ?
T2 = 38 C
WORK:
V1/T1= V2/T2
2.58 L = V2
288 K
= 2.79 L
311 K
C. Gay-Lussac’s Law
Temperature
(K)
Pressure
(torr)
P/T
(torr/K)
248
273
298
373
691.6
760.0
828.4
1,041.2
2.79
2.78
2.78
2.79
P
k
T
P
T
C. Gay-Lussac’s Law
 The
pressure and absolute
temperature (K) of a gas
are directly related
• at constant mass &
volume
P1/T1= P2/T2
P
T
D. Combined Gas Law
P
V
PV
PV = k
T
P 1V 1 P 2V 2
=
T1
T2
P 1 V 1T 2 = P 2V 2 T 1
E. Gas Law Problems
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P T WORK:
P1 = 765 torr
P1T2 = P2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K)
P2 = 560. torr
T2 = 217 K = -56°C
T2 = ?
E. Gas Law Problems
gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
A
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
T2 = 273 K
Gas Law Problems
gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
A
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
F. The First 4 Gas Laws
 The
Gas Laws Table