The Ideal Gas Law
Bringing It All Together
Objectives
When you complete this presentation, you will be able to
state the ideal gas law
derive the ideal gas law constant and discuss its units
use the ideal gas law to calculate pressure, volume,
temperature, or amount of gas in a system
calculate molar mass or density of a gas
Introduction
When we use the combined gas law
we allow P, V, and T to vary.
we keep the amount of gas constant.
If we vary the amount of gas as well as the pressure,
volume, and temperature
we will use the Ideal Gas Law
Introduction
When we use the phrase “Ideal Gas Law,”
we are talking about an ideal gas
we are not talking about a real gas
An ideal gas obeys all of the assumptions of the kinetic
theory of gases small particles
no attraction
moving rapidly
perfectly elastic collisions
Introduction
Under most common circumstances, real gases act like
ideal gases.
Only under conditions of low temperature
or
high pressure
will real gases deviate from ideality.
Application
Applying the kinetic theory, as we add an amount of gas
to a container of gas, we are introducing additional
particles to collide with the walls of the container.
Pressure goes up to keep V & T constant
Volume goes up to keep P & T constant
Temperature goes down to keep P & V constant
Application
This means that the amount (number of mols) of
material varies directly with pressure and volume
inversely with temperature
We will use the equation PV = nRT
P is pressure, V is volume, n is mols of gas, T is
temperature, and R is the gas constant.
Application
PV = nRT
P: measured in atm, kPa, or mm Hg
V: measured in L
n: measured in mol
T: measured in K
R: a constant whose value depends on the units of P,
V, n, and T
The Value of R
R is a constant whose value depends on the units of
P, V, n, and T (mostly on the units of pressure)
We can find R by solving the ideal gas law for R and
entering the appropriate values.
We remember from our studies of stoichiometry that
1.00 mol of a gas has a volume of 22.4 L at STP
(standard temperature and pressure).
This gives us enough information to find R for each
unit of pressure.
The Value of R
Where P is in atm:
STP is 1.00 atm at 273 K
pV = nRT ⇒ R = PV/nT
R = [(1.00 atm)(22.4 L)]/[(1.00 mol)(273 K)]
R = 0.0821 L-atm/mol-K
The Value of R
Where P is in mm Hg:
STP is 760 mm Hg at 273 K
pV = nRT ⇒ R = PV/nT
R = [(760 mm Hg)(22.4 L)]/[(1.00 mol)(273 K)]
R = 62.4 L-mm Hg/mol-K
The Value of R
Where P is in kPa:
STP is 101.3 kPa at 273 K
pV = nRT ⇒ R = PV/nT
R = [(101.3 kPa)(22.4 L)]/[(1.00 mol)(273 K)]
R = 8.314 L-mm Hg/mol-K
The Value of R
PV = nRT
If P is measured in atm:
R = 0.0821 atm-L/mol-K
If P is measured in kPa:
R = 8.314 kPa-L/mol-K
If P is measured in mm Hg:
R = 62.4 mm Hg-L/mol-K
The Value of R
PV = nRT
Remember:
The value of R is dependent on the units of pressure.
Always use the correct value of R.
All appropriate values for R will be given to you for any
test or quiz.
Example 1 – Finding P
What is the pressure, in atm, of 0.125 mols of helium in a
4.00 L container at a temperature of 430 K?
P = ? atm
V = 4.00 L
n = 0.125 mol
R = 0.0821 L-atm/mol-K
T = 430 K
PV = nRT
This is the value we
use when pressure is
expressed in atm.
nRT
(0.125)(0.0821)(430)
=
atm
⇒ P=
(4.00)
V
P = 1.10321875 atm = 1.10 atm
Practice Problems – Finding P
1. What is the pressure, in atm, of 3.50 mols of N2 gas held
in 45.0 L at a temperature of 310 K?
P = 0.198 atm
2. What is the pressure, in mm Hg, of 0.0400 mols of CO2
gas held in 10.0 L at a temperature of 350 K?
P = 87.4 mm Hg
3. What is the pressure, in kPa, of 172 mols of He gas held
in 675 L at a temperature of 273 K?
P = 578 kPa
4. What is the pressure, in atm, of 0.00250 mols of Cl2 gas
held in 0.100 L at a temperature of 455 K?
P = 0.934 atm
Example 2 – Finding V
What is the volume of 2.50 mols of oxygen at a pressure of
85.0 kPa and a temperature of 315 K?
P = 85.0 kPa
V=?L
n = 2.50 mol
R = 8.31 L-kPa/mol-K
T = 315 K
PV = nRT
nRT
=
⇒ V=
P
P = 76.98970588 L = 77.0 L
This is the value we
use when pressure is
expressed in kPa.
(2.50)(8.31)(315)
(85.0)
L
Practice Problems – Finding V
1. What is the volume of 1.00 mols of F2 gas at a pressure of
0.450 atm and a temperature of 298 K?
V = 54.4 L
2. What is the volume of 40.2 mols of UF6 gas at a pressure
of 645 mm Hg and a temperature of 655 K?
V = 2,550 L
3. What is the volume of 0.0424 mols of Ne gas at a pressure
of 4.53 kPa and a temperature of 242 K?
V = 18.8 L
4. What is the volume of 3.22 mols of O2 gas at a pressure
of 4.67 atm a temperature of 273 K?
V = 15.5 L
Example 3 – Finding n
How many mols of nitrogen is contained in a volume of 22.4 L
at a pressure of 760 mm Hg and a temperature of 273 K?
P = 760 mm Hg
V = 22.4 L
n = ? mol
R = 62.4 L-mm Hg/mol-K
T = 273 K
PV = nRT
PV
⇒ n=
RT
This is the value we
use when pressure is
expressed in mm Hg.
(760)(22.4)
= (62.4)(273) mol
P = 0.999342538 mol = 0.999 mol
Practice Problems – Finding n
1. How many mols of CO2 gas is in 75.0 L at a pressure of
2.75 atm and a temperature of 298 K?
n = 8.43 mol
2. How many mols of SF6 gas is in 0.500 L at a pressure of
950 mm Hg and a temperature of 350 K?
n = 0.0217 mol
3. How many mols of Ar gas is in 22.4 L at a pressure of
95.9 kPa and a temperature of 298 K?
n = 0.867 mol
4. How many mols of CH4 gas is in 0.0782 L at a pressure of
32.5 atm a temperature of 653 K?
n = 0.0474 mol
Example 4 – Finding T
What is the temperature of 1.60 mols of neon contained in
a volume of 15.0 L at a pressure of 1.20 atm?
P = 1.20 atm
V = 15.0 L
n = 1.60 mol
R = 0.0821 L-atm/mol-K
T=?K
PV = nRT
PV
⇒ T=
nR
This is the value we
use when pressure is
expressed in atm.
(1.20)(15.0)
= (1.60)(0.0821) K
P = 137.0280146 K = 137 K
Practice Problems – Finding T
1. What is the temperature of 3.00 mol of CO2 gas in 75.0 L
and at a pressure of 1.00 atm?
T = 305 K
2. What is the temperature of 0.755 mol of He gas in 4.25 L
and at a pressure of 2,320 mm Hg?
T = 209 K
3. What is the temperature of 35.0 mol of CH4 gas in 33.5 L
and at a pressure of 2,520 kPa?
T = 290 K
4. What is the temperature of 1.25 mol of CO2 gas in 25.0 L
and at a pressure of 1.70 atm?
T = 414 K
Application
The ability to measure the amount of the gas from
pressure, volume, and temperature measurements is a
powerful tool for exploring other properties of gases.
If we can also measure the mass of the gas, we can
determine
the molar mass of the gas
the density of the gas
The Molar Mass of a Gas
We can determine the number of mols, n, of a gas by
using pressure, volume, and temperature measurements
and the ideal gas law.
n = PV/RT
The molar mass, M, is the mass, m, divided by the
number of mols.
M =m/n
Putting the two equations together
M = m/(PV/RT) = mRT/PV
Example 5 – Finding M
At 301 K and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g.
What is the molar mass of this gas?
P = 0.974 atm
V = 1.00 L
R = 0.0821 L-atm/mol-K
T = 301 K
m = 5.16 g
M=
This is the value we
use when pressure is
expressed in atm.
mRT
(5.16)(0.0821)(301)
g/mol
=
(0.974)(1.00)
PV
M = 130.9183121 g/mol = 131 g/mol
Practice Problems – Finding M
1. At 302 K and 1.05 atm, 1.81 L of a gas has a mass of 5.42
g. What is the molar mass of this gas?
M = 70.0 g/mol
2. At 260 K and 695 mm Hg, 5.41 L of a gas has a mass of
10.2 g. What is the molar mass of this gas?
M = 44.0 g/mol
3. At 285 K and 97.2 kPa, 95.6 L of a gas has a mass of 329
g. What is the molar mass of this gas?
M = 83.9 g/mol
4. At 310 K and 4.15 atm, 0.350 L of a gas has a mass of
3.32 g. What is the molar mass of this gas?
M = 58.2 g/mol
The Density of a Gas
The density of a gas is the mass of the gas divided by its
volume
ρ = m/V
If we measure the density of a gas and know its
pressure and temperature, we can find the molar
mass, M .
mRT
m RT
m RT
ρRT
M=
PV = V P = V P = P
We separate
outWe replace
We are just rearranging
V
m/V with ρ .
the m/V term.
and P in the denominator.
Example 6 – Finding M
What is the molar mass of a gas with a density of 3.42 g/L at
293 K and 0.980 atm?
ρ = 3.42 g/L
R = 0.0821 L-atm/mol-K
T = 293 K
P = 0.980 atm
M=
This is the value we
use when pressure is
expressed in atm.
(3.42)(0.0821)(293)
ρRT
g/mol
=
(0.980)
P
M = 83.94808776 g/mol = 83.9 g/mol
Practice Problems – Finding M
1. What is the molar mass of a gas with a density of 3.25 g/L
at 150 K and 10.0 atm?
M = 4.00 g/mol
2. What is the molar mass of a gas with a density of 5.76 g/L
at 273 K and 672 mm Hg?
M = 146 g/mol
3. What is the molar mass of a gas with a density of 5.17 g/L
at 452 K and 672 kPa?
M = 352 g/mol
4. What is the molar mass of a gas with a density of 0.508
g/L at 345 K and 0.450 atm?
M = 32.0 g/mol
The Density of a Gas
If we know the molar mass of a gas, then we can
calculate the density of that gas under specific conditions
of pressure and temperature.
ρRT
M=
P
ρ=
MP
RT
wethis equation to
We If
used
rearrange
solve forand
molar mass
solve
… density.
whenfor
weρ knew
Example 7 – Finding ρ
What is the density of a sample of ammonia gas, NH3,
M = 17.04 g/mol, at 0.928 atm and 336 K?
M = 17.04 g/mol
P = 0.928 atm
R = 0.0821 L-atm/mol-K
T = 336 K
MP
ρ=
RT
(17.04)(0.928)
= (0.0821)(336) g/L
ρ = 0.5732382112 g/L = 0.573 g/L
This is the value we
use when pressure is
expressed in atm.
Practice Problems – Finding r
1. What is the density of a sample of CH4, M = 16.05 g/mol,
at 1.25 atm and 280 K?
r = 1.63 g/L
2. What is the density of a sample of H2, M = 2.02 g/mol, at
672 mm Hg and 261 K?
r = 0.0833 g/L
3. What is the density of a sample of CO2, M = 44.01 g/mol,
at 175 kPa and 310 K?
r = 3.00 g/L
4. What is the density of a sample of UF6, M = 352.0 g/mol,
at 10.2 atm and 397 K?
r = 110. g/L
Summary
The ideal gas law allows us to calculate the pressures,
volumes, amounts, and temperatures of gases.
PV = nRT
The value of the ideal gas constant, R, depends on the units of
pressure.
For units of atm, R = 0.0821 L-atm/mol-K
For units of kPa, R = 8.314 L-kPa/mol-K
For units of mm Hg, R = 62.4 L-mm Hg/mol-K
We can also use the ideal gas law to calculate the molar
masses and densities of gases.