Transcript Gas Laws - The Woodlands College Park High School

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
Definition/Relationship
Unit
Symbol
Pascal
Pa
SI pressure unit
1 Pa = Newton/meter2
Millimeters of
mercury
mmHg
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
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
• Gases are fluid – they flow
• Gases have low density
• Gases are compressible
• Gases 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 = V 2
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
Ideal Gas Law
• 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
Ideal Gas Law
• 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