Section 11–3:
The Ideal Gas Law
Coach Kelsoe
Chemistry
Pages 383–385
Section 11–3 Objectives
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State the ideal gas law.
Derive the ideal gas constant and discuss its
units.
Using the ideal gas law, calculate pressure,
volume, temperature, or amount of gas when the
other three quantities are known.
Using the ideal gas law, calculate the molar
mass or density of a gas.
Reduce the ideal gas law to Boyle’s law,
Charles’s law, and Avogadro’s law. Describe the
conditions under which each applies.
The Ideal Gas Law
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In Chapter 10 we talked about the three
quantities needed to describe a gas sample:
pressure, volume, and temperature.
We can further characterize a gas sample by
using a fourth quantity – the number of moles of
the gas sample.
The number of molecules or moles present will
always affect at least one of the other three
quantities.
The Ideal Gas Law
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Gas pressure, volume, temperature, and the
number of moles are all interrelated. There is a
mathematical relationship that describes the
behavior of a gas sample for any combination of
these conditions.
The ideal gas law is the mathematical
relationship among pressure, volume,
temperature, and the number of moles of a gas.
The equation for the ideal gas law is PV = nRT.
The Ideal Gas Law
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PV = nRT, where
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P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles of gas
R = the ideal gas constant
T = temperature (in Kelvins)
Since all these values are proportional, you could
actually derive the ideal gas law from Boyle’s
law, Charles’s law, and Avogadro’s law.
The Ideal Gas Constant
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In the equation representing the ideal gas law,
the constant R is known as the ideal gas
constant.
The value of R depends on the units chosen for
pressure, volume, and temperature.
For our calculations, we will let R be 0.0821
L·atm/mol·K. This means our measurements for
pressure must be in atm, our volume must be in
L, and our temperature must be in K.
The Ideal Gas Constant
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We can solve for R by doing R = PV/nT at STP:
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P = 1 atm
V = 22.41410 L
n = 1 mol
T = 273.15 K
R = PV/nT = (1 atm)(22.41410 L)/(1 mol)(273.15 K)
R = 0.08205784 L·atm/mol·K
The Ideal Gas Constant
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There are other values for R when you use
different units.
R = PV/nT at STP:
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P = 760 mm Hg
 V = 22.41410 L
 n = 1 mol
 T = 273.15 K
R = PV/nT = (760 mm Hg)(22.41410 L)/(1 mol)(273.15 K)
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R = 62.4 L·mm Hg/mol·K
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The Ideal Gas Constant
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There are other values for R when you use
different units.
R = PV/nT at STP:
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P = 101.325 kPa
 V = 22.41410 L
 n = 1 mol
 T = 273.15 K
R = PV/nT = (101.325 kPa)(22.41410 L)/(1 mol)(273.15 K)
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R = 8.314 L·kPa/mol·K
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Finding P, V, T, or n From the
Ideal Gas Law
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You can easily solve for pressure, volume,
temperature, or number of moles if you have
three of the other values.
The value for R is always the same, so you’ll
never have to solve for it.
P = nRT/V
V = nRT/P
n = PV/RT
T = PV/nR
Sample Problem 11–3
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What is the pressure in atmospheres exerted by
a 0.500 mol sample of nitrogen gas in a 10.0 L
container at 298 K?
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Given: V = 10.0 L, n = 0.500 mol, T = 298 K
Unknown: P of N2 in atmospheres
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PV = nRT  P = nRT/V
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P = (0.500 mol)(0.0821 L·atm/mol·K)(298 K) / 10.0 L
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P = 1.22 atm
Sample Problem 11–4
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What is the volume, in liters, of 0.250 mol of
oxygen gas at 20.0°C and 0.974 atm pressure?
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Given: P = 0.974 atm, n = 0.250 mol, T = 293.15 K
Unknown: V of O2 in liters
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PV = nRT  V = nRT/P
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V = (0.250 mol)(0.0821 L·atm/mol·K)(293.15 K) / 0.974
atm
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V = 6.17 L
Sample Problem 11–5
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What mass of chlorine gas, Cl2, in grams, is
contained in a 10.0 L tank at 27°C and 3.50 atm
of pressure?
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Given: P = 3.50 atm, V = 10.0 L, T = 300 K
Unknown: mass of Cl2 in grams
We don’t have a formula for mass, but we can
find it by finding the value of n.
PV = nRT  n = PV/RT
n = (3.50 atm)(10.0 L)/(0.0821
L·atm/mol·K)(300K)
n = 1.42 mol Cl2; 1.42 mol x 70.90 g/1 mol = 101 g
Finding Molar Mass from the
Ideal Gas Law
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You can solve for molar mass once you’ve found
the number of moles, but scientists have derived
another form of the ideal gas law to save us time.
Since the number of moles (n) is equal to mass
(m) divided by molar mass (M), we can substitute
m/M for the letter n.
We can use PVM = mRT, where P is pressure, V
is volume, M is molar mass (from periodic table),
m is the sample mass, R is the ideal gas
constant, and T is temperature.
Sample Problem 11–6
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At 28°C and 0.974 atm, 1.00 L of gas has a
mass of 5.16 g. What is the molar mass of this
gas?
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Given: P = 0.974 atm, V = 1.00 L, T = 301 K, m = 5.16 g
Unknown: M of gas in g/mol
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PVM = mRT  M = mRT/PV
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M = (5.16 g)(0.0821 L·atm/mol·K)(301 K)/(0.974 atm)(1.00 L)
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M = 131 g/mol
Finding Density from the Ideal
Gas Law
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Density (D) is equal to mass (m) divided by (V),
so therefore V = m/D. By subbing m/D for V, you
can solve for density if you are given molar
mass, and vice versa.
MP = DRT
Vocabulary
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Ideal gas constant
Ideal gas law