Transcript Gas Laws Chapter 14 - New Castle High School

Gas Laws
Chapter 14
Properties of Gases
 Gases
are easily compressed because
of the space between the particles in
the gas.
Properties of Gases
 The
amount of gas, volume, and
temperature affect the pressure of a
gas.
Properties of Gases
 Doubling
the number of particles in the
container would double the pressure
on a contained gas at constant
temperature.
 Boyle’s
Law
P1V1 = P2V2
 A)
50 kPa
 B) 33 kPa
 C) inverse proportion - as one
increases the other decreases
 Reducing
the volume of a contained
gas to one third, while holding
temperature constant, causes pressure
to become tripled.






A gas occupies a volume of 2.50 L at a
pressure of 350.0 kPa. If the temperature
remains constant, what volume would the
gas occupy at 1750 kPa?
V V2 = ? V1 = 2.50 L P1 = 350.0 kPa
P2 = 1750 kPa
F P1V1 = P2V2
S (350.0 kPa)(2.50 L) = (1750 kPa)V2
A V2 = 0.500 L
Boyles
 If
Law & Temperature
.
14B
the volume of a container of gas is
reduced, the pressure inside the
container will increase.
 The
graph of several pressure-volume
readings on a contained gas at
constant temperature would be a
curved line.
 If
a balloon is squeezed, the pressure
of the gas inside the balloon increases.
 Temperature
is directly proportional to
the average kinetic energy of the
particles in a substance.
 As
the temperature of the gas in a
balloon decreases, the average kinetic
energy of the gas decreases.
 Absolute
zero is the temperature at
which the average kinetic energy of
particles would theoretically be zero.
 This
is the lowest possible
temperature. (-273.15°C)
 To
get kelvins add 273 to the °C.
 To get °C subtract 273 from the kelvins.
A
temperature of -25°C is equivalent to
248 K. [ -25+273=248]
 A temperature of 295 K is equivalent to
22 °C. [295-273=22C]
 When
the Kelvin temperature of an
enclosed gas doubles, the particles of
the gas move faster.
 The
Kelvin temperature must be used
when working with proportions.
 Problem:
A
balloon contains 30.0 L of helium gas
at 103 kPa. What is the volume of the
helium when the balloon rises to an
altitude where the temperature stays
the same but the pressure is only 25.0
kPa?
V
V2 = ?
• P1 =103 kPa
• P2 = 25.0 kPa
F
S
A
V1 = 30.0 L
V2 = ?
P1V1 = P2V2
(103 kPa)(30.0 L) = (25.0 kPa)V2
V2 = 124 L
Charles’ Law
Notes 14C

V1
V2

T2
 T1
when the temperature is
expressed in kelvins.

At constant pressure, the volume of a
fixed mass of gas and its Kelvin
temperature are said to be directly
related.

If a balloon is heated, the
volume of the air in the balloon
increases if the pressure is
constant.
 The
temperature of 6.24 L of a gas is
increased from 125 K to 250 k at
constant pressure. What is the new
volume of the gas?
V

F

V2 =? V1 = 6.24 L
T1 = 125 K T2 = 250. K
V1
V
 2
T1
T2

V2
6.24 L
=
125 K 250. K
A
V2 = 12.48 L
S
 A)
kelvins
B) increases C) 0 L
Combined Gas Law
Notes 14 D

Gay-Lussac’s Law
P1 P2

T1 T2
 If
the Kelvin temperature of a gas in a
closed container increases the
pressure of the gas increases
proportionally.
 As
the temperature of a fixed volume of
a gas increases, the pressure will
increase.
A
sample of chlorine gas has a
pressure of 9.99 kPa at 27°C. What will
its pressure be at 627°C if its volume
remains constant?
V
P2 = ? P1 = 9.99 kPa
 T1 = 300. K T2 = 900. K
F
P1 P2


T1
S
P2
9.99 k Pa
=
300. K
900 K

A
T2
P2 = 29.97 kPa
 The
Combined Gas Law
P1 V1
P2 V2

T1
T2
The combined gas law relates
temperature, pressure, and volume.
 If
a sample of oxygen occupies a
volume of 6.00 L at a pressure of 68.0
kPa and a temperature of 264 K, what
volume would this sample occupy at
204 kPa and 528 K?
V


F
S
V2 =?
P1 = 68.0 kPa
P2 = 204 kPa
P1 V1 P2 V2

T1
T2
(68.0 kPa)(6.00 L) (204 kPa)V2
264 K
A
V1 = 6.00 L
T1 = 264 K
T2 = 528 K
V2 = 4.00 L
=
528 K
Ideal Gas Law
Notes 14E
 Ideal
Gas Law PV = nRT
 Where n = the number of moles
L  kPa
R is the Ideal Gas Constant R  8.31
K  mol
 The
ideal gas law can be used to
calculate the number of moles of a
contained gas.
 What
is the volume (in L) that would be
occupied by 1.00 mol of O2 at STP?
 V V = ? n = 1.00 mol T = 273K
L  kPa
P = 101.3 kPa
R  8.31
K  mol
F
PV = nRT
 S (101.3 kPa)V = (1.00 mol)
A
Vol = 22.4 L
R  8.31
L  kPa
(273 K)
K  mol
 How
many moles of H2 would be
contained in 83.1 L of the gas at
137 kPa and 1.0°C?
V
n=?
V = 83.1 L
P = 137 kPa
T = 1.0°C = 274 K R  8.31 L  kPa
K  mol

F
PV = nRT
 S (137 kPa)(83.1 L) = n
 A n = 5.00 mol
 L  kPa 
 8.31

 K  mol 
(274. K)
Gas Mixtures
14 F
 Dalton’s
Law - In a mixture of gases,
the total pressure is the sum of the
partial pressures of the gases.
 PTotal
= P1 + P2 + P3 + …
A
sample of H2 is collected over water
such that the combined hydrogenwater vapor sample is held at a
pressure of 1 standard atmosphere.
What is the partial pressure of the H2 if
that of the water vapor is 2.5 kPa?

1 atm = 101.3 kPa
 101.3
kPa – 2.5 kPa = 98.8 kPa
 The
tendency of molecules to move
toward areas of lower concentration is
called diffusion.
 The gas propellant in an aerosol can
moves from a region of high pressure
to a region of lower pressure.

The process that occurs when a gas
escapes through a tiny hole in the
container is called effusion.
The substance with the smallest
molar mass would have the fastest
rate of effusion.
 So CH4 effuses faster than NO2.

16g/mol
46g/mol

