Transcript Gas Laws - Valley Mills Independent School District

Gas Laws
Kinetic Theory
• Particles of matter are ALWAYS in
motion.
• The volume of individual particles is
approximately zero.
• Collision of particles with container wall
causes pressure.
• Particles exert no forces on each other.
• The average kinetic energy is
approximately equal to the Kelvin
temperature of gases.
Measuring Pressure
• The first device for
measuring
atmospheric pressure
was developed by
Evangelista Torricelli
during the 17th century
• The device was called
a “barometer”
– baro = weight
– meter = measure
The Early Barometer
• The normal
pressure due to
the atmosphere
at sea level can
support a column
of mercury that
is 760 mm high
Modern Day Barometers
Pressure Units
• Caused by the collisions of molecules
colliding with the wall of the container
• equal to force/unit area
• SI unit = Newton/meter2 = 1 pascal
(Pa)
• 1 standard atmosphere = 101.31 kPa
• 1 standard atmosphere = 1 atm = 760
mmHg = 760 torr
Pressure Units
(KNOW THIS CHART!!!!!)
Unit
Symbol
Definition/Relationship
Pascal
Pa
Millimeters
of mercury
mmHg
SI pressure unit
1 Pa = Newton/meter2
Pressure that supports a 1
millimeter column of
mercury in a barometer
Atmosphere
Atm
Torr
Torr
Average atmospheric
pressure at sea level
and 0° C
1 torr = 1 mmHg
Converting Celsius to Kelvin
• For every 1 °C you cool a gas,
the volume decreases by 1/273.
• –273 °C is absolute zero (0 K)
– the coldest temperature possible;
all motion ceases.
• K = °C + 273
• °C = K - 273
Converting Celsius to Fahrenheit
 F = 9/5  C + 32
 C = 5/9 ( F- 32)
Standard Temperature and Pressure
(STP)
• Pressure:
1 atm or 760 mmHg or 101.3
kPa
• Temperature:
°
0 C or 273 K
The Nature of Gases
Gases:
• expand to fill their
container
• are fluid – they flow
• have low density
• are compressible
• effuse and diffuse
Boyle’s Law
• Robert Boyle (1627-1691), Irish
chemist
• The volume of a given amount of gas
held at a constant temperature varies
inversely with the pressure. As
pressure increases, volume
decreases.
• P1V1 = P2V2
Boyle’s Law
• Solve the following:
825 Torr = _____ kPa
101.3 kPa / 760 Torr = x kPa / 825 Torr
109.96 kPa = x
• Solve the following:
The volume of oxygen at 120 kPa is 3.20 L.
What is the volume of oxygen at 101.3 kPa?
P1V1 = P2V2
(120 kPa)(3.20L) = (101.3 kPa)V2
3.79 L = V2
Charles’ Law
• Jacques Charles (1746-1823), French
physicist
• The volume of a gas varies directly with its
absolute temperature. As temperature
increases, volume increases.
• V1 = V2
T1
T2
Charles’ Law Cont.
• Solve the following:
– The volume of a gas at 40.0 °C is 4.50 L.
Find the volume of the gas at 80 °C .
40°C + 273 = 313 K 80°C + 273 = 353 K
V1/T1 = V2/T2
4.5L / 313 K = V2 / 353 K
9 L = V2
Gay-Lussac’s Law
• Joseph Gay-Lussac
• The pressure of a given mass of gas
varies directly with the Kelvin
temperature when the volume remains
constant.
• P1 = P2
T1
T2
*** Remember – temperature must be in
Kelvin
Gay-Lussac’s Law
• Solve the following:
– The pressure of a gas in a tank is 3.20 atm at
22.0 °C. If the temperature rises to 60.0 °C
what will be the gas pressure in the tank?
Temp must be in Kelvin, so convert °C to K
T1 = 22.0 °C + 273 = 295K T2 = 60°C + 273 = 333K
P1/T1 = P2/T2
3.2 atm / 295K = x atm / 333K
3.61 atm = P2
Combined Gas Law
• Boyle’s, Charles’s, and Gay Lussac’s laws
can be combined into a single law.
• The combined gas law states the
relationship among pressure, volume, and
temperature of a fixed amount of gas.
• P1V1 = P2V2
T1
T2
*** Remember – temperature must be in
Kelvin
Combined Gas Law
• Solve the following:
– A gas at 110 kPa and 30.0 °C fills a flexible
container with an initial volume of 2.00 L. If
the temperature is raised to 80.0 °C and the
pressure increased to 440 kPa, what is the new
volume? Convert temp to Kelvin:
T1 = 30°C + 273 = 303K
T2 = 80°C + 273 = 353K
P1V1/T1 = P2V2/T2
(110 kPa)(2.00L) / 303K = (440 kPa)V2 / 353K
0.582 L = V2
Ideal Gas Law
• In all the other gas laws, the relationships hold
true for a “fixed mass” or “given amount” of a gas
sample.
• Because pressure, volume, temperature, and the
number of moles present are all interrelated, one
equation is used to describe their relationship.
PV = nRT
• n = moles
• R = ideal gas law constant
• V must be in liter
• T must be in Kelvin
The value of R depends on
pressure unit given.
– If pressure unit is mm Hg orTorr :
• R = 62.4 L • mm Hg or Torr
mol • K
– If pressure is atm:
• R = 0.0821 L • atm
mol • K
– If pressure is kPa:
• R = 8.31 L • kPa
mol • K
Ideal Gas Law
• Solve the following:
44.01 g of CO2 occupies a certain volume
at
STP. Find that volume. n = sample mass
molar mass
n = 44.01 g
44.0098g/mol
n = 1 mol
(1atm)(V) = (1mol)(0.0821) (273 K)
V = 22.4 L
Solve the following:
Find the molar mass of a gas that
weights 0.7155 g and occupies a volume
of 250. mL at STP.
(1 atm)(0.250 L) = (n)(0.0821)(273 K)
n = 0.011 mol
0.011 mol = 0.755 g
X
X = 68.64 g/mol
Dalton’s Law of Partial Pressure
• The total pressure of a mixture of gases
equals the sum of the pressures that
each would exert if it were present alone.
• Pt is the total pressure of a sample which
contains a mixture of gases
• P1, P2, P3, etc. are the partial pressures
of the gases in the mixture
Pt = P1 + P2 + P3 + ...
• The pressure of a mixture of nitrogen,
carbon dioxide, and oxygen is 150 kPa.
What is the partial pressure of oxygen if the
partial pressures of the nitrogen and carbon
dioxide are 100 kPA and 24 kPa,
respectively?
Pt = P1 + P2 + P3 + ...
P = PN2 + PCO2 + PO2
150 kPa = 100 kPa + 24 kPa + PO2
PO2 = 150 kPa - 100 kPa - 24 kPa
PO2 = 26 kPa