Unit 13 – Gas Laws
Gases
The Gas Laws
Kinetic Theory Revisited
1. Particles are far apart and have negligible
volume.
2. Move in rapid, random, straight-line
motion.
3. Collide elastically.
4. No attractive or repulsive forces.
The Gas Laws
Variables That Describe A Gas
1. Pressure (P)
•
kPa (atm and mmHg)
2. Volume (V)
•
L (mL)
3. Temperature (T)
•
K
• TK = 273 + T ºC
4. Amount of gas (n)
•
mol
STP – Not Just a Motor Oil
• STP – standard temperature and pressure
– 0ºC (273 K)
– 1 atm (101.3 kPa)
The Gas Laws
Amount of a Gas
• As the number of particles increases the
pressure increases.
n
P
The Gas Laws
Volume
• As volume increases the pressure
decreases.
V
P
The Gas Laws
Temperature
• As the temperature increases the pressure
increases.
T
P
The Gas Laws
Boyle’s Law
• At constant temperature and mass (moles),
the pressure is inversely proportional to the
volume.
P1V1 = P2V2
Solving a gas law problems
1. Set up a variable table.
•
Are the units consistent and appropriate?
•
Consistent
•
•
Pressures and volumes must be in the same type of unit.
Appropriate
•
•
Temperature and amounts must be in Kelvin and moles.
Pressure and volumes might have to be in a specified unit.
2. Establish which equation you’ll be using.
3. Solve the equation for the unknown variable.
•
DO NOT SUBSTITUTE YET!!!
4. Substitute and solve!
Boyle’s Law Problem
A balloon contains 30.0 L of helium gas at 103 kPa. What is
the volume when rises to an altitude where the pressure is
only 25.0 kPa? (assume the temp is constant)
P1 = 103 kPa
V1 = 30.0 L
P2 = 25.0 kPa
V2 = ?
P1V1 = P2V2
P2
V2 =
P1V1
P2
=
P2
(103 kPa)(30.0 L)
25.0 kPa
= 123.6 L
124 L
The Gas Laws
Charles’s Law
• At constant pressure and mass (moles), the
volume of a gas is directly proportional to
the Kelvin temperature.
V1
T1
=
V2
T2
Charles’s Law Problem
A balloon inflated in a room at 24 °C has a volume of 4.00 L.
The balloon is then heated to a temperature of 58 °C. What is
the new volume if pressure is constant?
T1 = 24 °C 297 K T2 = 58 °C 331 K
V1 = 4.00 L
V2 = ?
Temperature MUST be in Kelvin (add 273)
T2 V1
V2 T2
=
T1
T2
T2V1
(331 K)(4.00 L)
V2 =
=
= 4.4579 L
T1
(297 K)
4.46 L
The Gas Laws
Gay-Lussac’s Law
• At constant volume and mass (moles), the
pressure of a gas is directly proportional to
the Kelvin temperature.
P1
T1
=
P2
T2
Gay-Lussac’s Law Problem
The gas left in a used aerosol can is at a pressure of 103 kPa
at 25 °C. If the can is thrown onto a fire and the pressure
increases to 415 kPa, what is the temperature?
P1 = 103 kPa
T1 = 25 °C 298 K
P1
P2 T1
=
P2 = 415 kPa
T2 = ?
P2
T2 P2
Do we want to solve for 1 ?
T2
Gay-Lussac’s Law Problem
The gas left in a used aerosol can is at a pressure of 103 kPa
at 25 °C. If the can is thrown onto a fire and the pressure
increases to 415 kPa, what is the temperature?
P1 = 103 kPa
T1 = 25 °C 298 K
T1T2 P1
P1 T1
=
P2 = 415 kPa
T2 = ?
P2 T2 T1
T2 P1
P2 T1
(415 kPa)(298 K)
T2 =
=
= 1200.6 K
P1
103 kPa
1.20  103 K
The Combined Gas Laws
Combined Gas Law
P1 V1
T1
=
P2 V2
T2
Boyle’s Law Lab Reminders
1. Watch rounding and units everywhere!
1. Data #1,3,4 and Calculations #1 need units.
2. Rounding and units for Data #3 and 4 are determined
by the data used to make your graph.
3. If you have the units correct, you actually should get
mmHg/B for Calculation #2 (B is book).
4. The final answer should be in the 1000’s.
5. Does your graph have appropriate labels everywhere
(title and axis) and units?
Where do all of these gas
laws come from?
What happened to moles?
The Ideal Gas Law
Ideal Gas Law
PV = nRT
P = pressure (kPa, atm, or mmHg)
V = volume (L)
n = moles (mol)
T = temperature (K)
R = ideal gas constant
R = 8.31
L•kPa
mol •K
or 0.0820
L•atm
mol •K
or 62.4
L•mmHg
mol •K
The Ideal Gas Law
ALL OF THE OTHER GAS LAWS CAN BE
DERIVED FROM THE IDEAL GAS
LAW!
Solve for the
PV = nRT
constant R!
nT nT
P1V1
n1T1
=R=
P2V2
n2T2
Ideal Gas Law Problem
What is the volume of 14.2 g of O2(g) at STP?
PV = nRT
P
14.2 g O2
V
P
1 mol O2
= .444 mol O2
32.0 g O2
(.444 mol)(8.31 )(273K)
=
101.3 kPa
L•kPa
mol •K
= 9.94 L
The Ideal Gas Law
• Other expressions of the Ideal Gas Law
– molar mass (M) is:
mass(m)
M
mole(n)
– Therefore n = m/M, substituting this into the ideal
gas law yields
mRT
PV 
M
The Ideal Gas Law
• Solving for molar mass (M)
mRT
M
PV
• Since density (D) = mass/volume
DRT
M
P
• Solving for density
MP
D
RT
The Ideal Gas Law
• Ideal Gases follow the assumptions of kinetic
theory.
– No attractive forces between particles and
negligible volume
– Under most temperatures and pressures most real
gases behave like ideal gases
• When do gases become less ideal and more real?
– Low temperature
– High pressure
Going Back particles
Avogadro’s Principle
Equal volumes of gases at the same
temperature and pressure contain an equal
number of particles.
Molar volume
• Solving for volume in the Ideal Gas Law for
1 mole at STP we obtain:
nRT
V
P
LkPa
molK
(1 mole)(8.314
)(273K )
V
 22.4 L
101.3kPa
Therefore, at STP 1 mole = 22.4 L
This known as the Standard Molar Volume
Remember this problem!
What is the volume of 14.2 g of O2(g) at STP?
(Same question we did in Mole Unit)
The previous equality gives us shortcut for
any question at STP!
14.2 g O2 1 mol O2 22.4 L O2
32.0 g O2 1 mol O2
= 9.94 L O2
This is the same answer as before!
Don’t Forget About Dalton!!!
• Dalton’s Law Correction
– Because most gases are COLLECTED OVER
WATER, you will often use this law to correct
the pressure.
Ptotal = Pgas + Pwater
Solving for the gas yields:
This is usually
the P1 or P in a
gas law
problem!
Pgas = Ptotal – Pwater