# The gas laws

Section 2 – The Gas Laws
Scientists have been studying physical properties of gases for
hundreds of years. In 1662, Robert Boyle discovered that gas
pressure and volume are related mathematically. The
observations of Boyle and others led to the development of
the gas laws. The gas laws are simple mathematical
relationships between the volume, temperature, pressure, and
amount of a gas.
The gas laws
Since gases are highly
compressible and will expand
when heated, these properties
have been studied extensively.
The relationships between volume,
pressure, temperature and moles
are referred to as the gas laws.
To understand the relationships, we
must look a few concepts.
5-2
Units we will be using
Volume
liters, although other units
could be used.
Temperature
Must use an absolute scale.
K - Kelvin is most often used.
Pressure
Atm, torr, mmHg, lb/in2.
- use what is appropriate.
Moles
We specify the amounts in
molar quantities.
5-3
Gas laws
Laws that show the relationship between
volume and various properties of gases
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
The Ideal Gas Equation combines several of
these laws into a single relationship.
5-4
Boyle’s law
The volume of a gas is inversely proportional
to its pressure.
PV = k
or
P1 V1 = P2 V2
Temperature and number of moles must be
held constant!
5-5


Boyle’s law can be used to compare
changing conditions for a gas
Using P1 and V1 to stand for initial
conditions and P2 and V2 to stand
for new conditions results in the
following equations
P1V1 =k
P2V2 = k
P1V1 = P2V2
Practice Problem
A sample of oxygen gas has a volume of 150. mL
when its pressure is 0.947 atm. What will the
volume of the gas be at a pressure of 0.987 atm if
the temperature remains constant?


Given: V1 of O2 = 150. mL; P1 of O2 = 0.947
atm; P2 of O2 = 0.987 atm
Unknown: V2 of O2 in mL
Given: V1 of O2 = 150. mL; P1 of O2 = 0.947 atm; P2 of O2
= 0.987 atm
Unknown: V2 of O2 in mL
Rearrange the equation for Boyle’s law
(P1V1 = P2V2) to obtain V2
Practice Problems
A balloon filled with helium gas has a volume of 500
mL at a pressure of 1 atm.The balloon is released
and reaches an altitude of 6.5 km, where the
pressure is 0.5 atm. Assuming that the temperature
has remained the same, what volume does the gas
occupy at this height?

A gas has a pressure of 1.26 atm and occupies a
volume of 7.40 L. If the gas is compressed to a
volume of 2.93 L, what will its pressure be, assuming
constant temperature?

Divers know that the pressure exerted by the water
increases about 100 kPa with every 10.2 m of depth.
This means that at 10.2 m below the surface, the
pressure is 201 kPa; at 20.4 m, the pressure is 301
kPa; and so forth. Given that the volume of a balloon is
3.5 L at STP and that the temperature of the water
remains the same, what is the volume 51 m below the
water’s surface?

Charles’s Law: Volume-Temperature

Charles’s law  the volume of a fixed mass of
gas at constant pressure varies directly with the
Kelvin temperature
Charles’ law
When you heat a sample of a gas, its volume increases.
The pressure and number
of moles must be held constant.
5 - 14
Charles’ Law
Placing an air filled balloon near liquid nitrogen (77 K)
will cause the volume to be reduced. Pressure and
the number of moles are constant.
5 - 15

The temperature −273.15&deg;C is referred to as
absolute zero and is given a value of zero in the
Kelvin scale
K = 273 + &deg;C
V = kT
or
V/T = k
Practice Problems
A sample of neon gas occupies a volume of 752
mL at 25&deg;C.What volume will the gas occupy at
50&deg;C if the pressure remains constant?
Given: V1 of Ne = 752 mL; T1 of Ne = 25&deg;C + 273
= 298 K; T2 of Ne = 50&deg;C + 273 = 323 K
 Note
that Celsius temperatures have been converted to
kelvins.This is a very important step for working the
problems in this chapter
Unknown: V2 of Ne in mL
Given: V1 of Ne = 752 mL; T1 of Ne = 25&deg;C + 273 = 298 K; T2 of Ne = 50&deg;C +
273 = 323 K
Unknown: V2 of Ne in mL
A helium-filled balloon has a volume of 2.75 L at
20.&deg;C.The volume of the balloon decreases to 2.46
L after it is placed outside on a cold day. What is
the outside temperature in K? in &deg;C?

A gas at 65&deg;C occupies 4.22 L. At what Celsius
temperature will the volume be 3.87 L, assuming the
same pressure?

Gay-Lussac’s Law: Pressure-Temperature

Gay-Lussac’s law: The pressure of
a fixed mass of gas at constant
volume varies directly with the Kelvin
temperature
P = kT
or
P/T = k
Gay-Lussac’s Law
Law of of Combining Volumes.
At constant temperature and
pressure, the volumes of gases
involved in a chemical reaction are in
the ratios of small whole numbers.
Studies by Joseph Gay-Lussac led to a
better understanding of molecules and
their reactions.
5 - 22
Gay-Lussac’s Law
Example.
Reaction of hydrogen and oxygen gases.
H2
H2
+
O2
H2O H2O
Two ‘volumes’ of hydrogen will combine
with one ‘volume’ of oxygen to produce two
volumes of water.
We now know that the equation is:
2 H2 (g) + O2 (g)
2 H2O (g)
5 - 23
The gas in an aerosol can is at a pressure of 3.00 atm at
25&deg;C. Directions on the can warn the user not to keep
the can in a place where the temperature exceeds
52&deg;C. What would the gas pressure in the can be at
52&deg;C?


Given: P1 of gas = 3.00 atm; T1 of gas = 25&deg;C +
273 = 298 K; T2 of gas = 52&deg;C + 273 = 325 K
Unknown: P2 of gas in atm
Given: P1 of gas = 3.00 atm; T1 of gas = 25&deg;C + 273 = 298 K; T2 of gas = 52&deg;C +
273 = 325 K
Unknown: P2 of gas in atm
Before a trip from New York to Boston, the pressure in
an automobile tire is 1.8 atm at 20.&deg;C. At the end
of the trip, the pressure gauge reads 1.9 atm. What
is the new Celsius temperature of the air inside the
tire? (Assume tires with constant volume.)

At 120.&deg;C, the pressure of a sample of nitrogen is
1.07 atm. What will the pressure be at 205&deg;C,
assuming constant volume?

A sample of helium gas has a pressure of 1.20 atm at
22&deg;C. At what Celsius temperature will the helium
reach a pressure of 2.00 atm?

Equal volumes of gas at the same
temperature and pressure contain equal
numbers of molecules.
V=kn
V1
V2
=
n1
n2
5 - 29
If you have more moles of a gas, it takes up more
space at the same temperature and pressure.
5 - 30
Standard conditions (STP)
Remember the following standard conditions.
Standard temperature
Standard pressure
= 273.15 K
= 1 atm
At these conditions:
One mole of a gas has
a volume of 22.4 liters.
5 - 31
The ideal gas law
A combination of Boyle’s, Charles’ and
PV = nRT
P = pressure, atm
V = volume, L
n = moles
T = temperature, K
R = 0.082 06 L atm/K mol
(gas law constant)
5 - 32
Example
What is the volume of 2.00 moles of gas at
3.50 atm and 310.0 K?
PV = nRT
V = nRT / P
= (2.00 mol)(0.08206 L atm K-1mol-1)(310.0 K)
(3.50 atm)
= 14.5 L
5 - 33
Ideal gas law
R can be determined from standard conditions.
R=
PV
nT
R=
( 1 atm ) ( 22.4 L )
( 1 mol ) ( 273.15 K)
= 0.08206 atm L mol-1 K-1
5 - 34
Ideal gas law
When you only allow volume and one other
factor to vary, you end up with one of the
other gas laws.
Just remember
Boyle
Charles
Pressure
Temperature
Moles
5 - 35
Ideal gas law
P1 V 1
n1T1
=R=
P2V2
n2T2
This one equation
says it all.
Anything held constant will
“cancels out” of the equation
5 - 36
Ideal gas law
Example - if n and T are held constant
P1V1
n1T1
=
P2V2
n2T2
This leaves us
P1V1 = P2V2
Boyle’s Law
5 - 37
Example
If a gas has a volume of 3.0 liters at 250 K,
what volume will it have at 450 K ?
Cancel P and n
They don’t change
We end up with
Charles’ Law
P1V1
n1 T 1
V1
T1
=
=
P2V2
n 2T2
V2
T2
5 - 38
Example
If a gas has a volume of 3.0 liters at 250 K,
what volume will it have at 450 K ?
P1V1
n1 T 1
=
P2V2
V1
n2T 2
T1
V2
=
=
(3.0 l) (450 K)
(250 K)
=
5.4 L
V2
T2
5 - 39
Dalton’s law of
partial pressures
The total pressure of a gaseous mixture is the
sum of the partial pressure of all the gases.
PT
=
P1 + P2 + P3 + .....
Air is a mixture of gases - each adds it own
pressure to the total.
Pair = PN2 + PO2 + PAr + PCO2 + PH2O
5 - 40
Partial pressure example
Mixtures of helium and oxygen are used in
scuba diving tanks to help prevent “the
bends.”
For a particular dive, 46 liters of O2 and 12
liters of He were pumped in to a 5 liter tank.
Both gases were added at 1.0 atm pressure
at 25oC.
Determine the partial pressure for both gases
in the scuba tank at 25oC.
5 - 41
Partial pressure example
First calculate the number of moles of each
gas using PV = nRT.
nO 2
nHe
(1.0 atm) (46 l)
= (0.08206 l atm K-1 mol-1)(298.15K) = 1.9 mol
(1.0 atm) (12 l)
= (0.08206 l atm K-1 mol-1)(298.15K) = 0.49 mol
5 - 42
Partial pressure example
Now calculate the partial pressures of each.
PO2 =
(1.9 mol) (298.15 K) (0.08206 l atm K-1 mol-1)
(5.0 l)
= 9.3 atm
PHe =
(0.49 mol) (298.15 K) (0.08206 l atm K-1 mol-1)
(5.0 l)
= 2.4 atm
Total pressure in the tank is 11.7 atm.
5 - 43
Diffusion
Diffusion and effusion
Diffusion
The random and
spontaneous mixing of
molecules.
Effusion
The escape of
molecules through
small holes in a
barrier.
5 - 45
Practice Problems
A helium-filled balloon has a volume of 50.0 L at
25&deg;C and 1.08 atm.What volume will it have at
0.855 atm and 10.&deg;C?
Given: V1 of He = 50.0 L; T1 of He = 25&deg;C + 273 =
298 K; T2 of He = 10&deg;C + 273 = 283 K; P1 of He
= 1.08 atm; P2 of He = 0.855 atm
Unknown: V2 of He in L
Given: V1 of He = 50.0 L; T1 of He = 25&deg;C + 273 = 298 K; T2 of He =
10&deg;C + 273 = 283 K; P1 of He = 1.08 atm; P2 of He = 0.855 atm
Unknown: V2 of He in L
The volume of a gas is 27.5 mL at 22.0&deg;C and 0.974
atm. What will the volume be at 15.0&deg;C and 0.993
atm?

A 700. mL gas sample at STP is compressed to a
volume of 200. mL, and the temperature is
increased to 30.0&deg;C. What is the new pressure of
the gas in Pa?

Answer 3.94 &times; 105 Pa, or 394 kPa

Given: PT = Patm = 731.0 torr; PH2O = 17.5 torr
(vapor pressure of water at 20.0&deg;C); Patm = PO2 +
PH2O
Unknown: PO2 in torr
PO2 = Patm − PH2O
PO2 = 731.0 torr − 17.5 torr = 713.5 torr
Some hydrogen gas is collected over water at
20.0&deg;C.The levels of water inside and outside the
gas-collection bottle are the same. The partial
pressure of hydrogen is 742.5 torr. What is the
barometric pressure at the time the gas is collected?
