Ideal Gas Law
The previous gas laws allows us to calculate volume, pressure and temperature, but not an
amount of gas in a container.
There is an equation that will allow us to determine the amount of gas it is called the Ideal Gas
Law and the mathematical expression is: PV=nRT. You should recognize 3 of the 5
variables P is the pressure, V is the volume and T is the temperature. n represents the moles of
the gas.
What is the “R”? It is called the Ideal Gas constant. The equation you are using is for an ideal
gas which conforms to all of the Kinetic Theory Assumptions, but we are using this equation
for Real Gases which do not conform to all of the assumptions. The 2 assumptions that real
gases do not conform to are:
1. Real Gases do have some attractions between their particles.
2. Real Gases have a volume.
But Real Gases do act closely like Ideal gases except at low temperatures and high
pressures.
So “R” regulates the Real gases for the Ideal Gas Law Equation.
Now R has a value of:
8.31 (L) (kPa)
or
(moles) (K)
.0821 (L) (atm)
(moles) (K)
or 62.4 (L) (mm Hg)
(moles) (K)
The reason “R” has different values is because of the Pressure units. YOU MUST KNOW
ALL 3 VALUES FOR “R”. Also notice the volume unit is always in LITERS.
But the Ideal Gas Law now allows us to determine the moles of a gas and consequently grams
of a gas. I will demonstrate 3 examples for you and using these examples, you are to complete:
Pg 465 #26&27, Pg 466 #28 & 29, and Pg 480 #’s 65-69. If you need to, finish for
homework.!!
Example 1: At 34oC, the pressure inside a nitrogen filled tennis ball with a volume of 0.148L is
212kPa. How many moles of N2 gas are in the tennis ball?
Using PV=nRT, we need to get “n” alone so
PV = n P = 212kPa, V = 0.148L,
RT
R = 8.31 (L) (kPa) T = 34oC + 273 =
307K
(moles) (K)
(212kPa)(0.148L)___
8.31 (L) (kPa) (307K)
(moles) (K)
= 0.0123 moles of N2 gas
Example 2: A deep underground cavern contains 2.24 X 106 L of CH4 gas at a pressure of
1.50 x 103 kPa and a temperature of 315K. How many grams of CH4 does this
cavern contain?
Note: I will find moles first using the Ideal Gas Law and then convert the moles to grams.
P = 1.50 x 103 kPa
V = 2.24 X 106 L
T = 315K R = 8.31 (L) (kPa)
(moles) (K)
PV = n
RT
(1.50 x 103kPa)( 2.24 X 106 L )___
8.31 (L) (kPa) (315K)
= 1.28 X 106 moles of CH4
(moles) (K)
Now to convert to grams: 1.28 X 106 moles x 16.0gram = 2.05 x 107 grams CH4
1 mole
Example 3: What is the pressure of a gas in kPa, of 20.0 grams of Argon at a temperature of
92oC and contains a volume of 235,000ml?
We will use PV = nRT to find the pressure but you must complete a few steps first.
1. I need to change my grams of Argon to moles: 20.0 grams x
1mole = 0.501moles
39.9 grams
2. I need to change my volume units to Liters because the “R” constant unit is in Liters.
235,000ml x
0.001L = 235 L
1ml
Now I could use PV = nRT. First get P alone: P = nRT
V
o
n = 0.501 moles, R = 8.31 (L) (kPa) , T = 92 C + 273 = 365K
V = 235L
(moles) (K)
(0.501 moles) 8.31 (L) (kPa) (365K)
_________
(moles) (K)
______________ = 6.47 kPa
235L
All 3 example I used the same “R” value, but remember there are 2 “R” values your are
responsible for and may have to use in the problems assigned.