April 30, 2013
Positive Attitude
Peccadillo: a slight offense or fault; a minor sin
• Do Now: Quad Card
Topic: Behavior of Gases
The Ideal Gas Law
The Ideal Gas Law
• remember, the gas laws are simple, mathematical relationships
between the pressure (P), volume (V), temperature (T), and
moles (n), of a gas.
• so far, we have only dealt with P, V, and T (in the basic gas laws).
• today, we will introduce n, the number of moles of a substance.
• remember, moles are related to all sorts of good stuff, like
grams and molecules!
• “n,” however, must always be expressed in moles. This
means if you are given grams or molecules, you must convert
them into moles before using it in the Ideal Gas Law!
• so , what is the Ideal Gas Law? Here it is:
PV = nRT
• “R” is known as the ideal gas law constant.
• its value never changes (duh).
R = .0821 Latm
• notice the unit for R. It’s complicated!
molK
The Ideal Gas Law
• for the Ideal Gas Law ONLY, all units for P, V,
R = .0821 Latm
n, and T must match the units in the R value.
molK
• this means your P must be in atm, V must
be in L, n must be in mol, and of course, T must be in K.
• if they don’t match, convert them!
• this is ONLY for the Ideal Gas Law! You may still use kPa,
mmHg, mL, etc. in the Basic gas laws, just not in the Ideal!
• therefore, you should expect to be doing a lot of conversions
when doing Ideal Gas Law calculations!
Ex1: If 234.68 g of Cl2 was compressed at 3534 mmHg of pressure and 13.8C, what volume would it have? (1 mol Cl2 = 70.90 g)
The Ideal Gas Law
Ex1: If 234.68 g of Cl2 was compressed at 3534 mmHg of pressure and
-13.8C, what volume would it have?
= 4.65 atm
P = 3534 mmHg _____
4.65 atm 3534 mmHg 1 atm
760 mmHg
V=
?
3.31 mol
n = 234.68 g  _____
234.68 g Cl2 1 mol Cl2 = 3.31 mol Cl2
259.2 K
T = -13.8 C  ______
70.90 g Cl2
PV
nRT
__ = ___
P
P
V = nRT
.0821 Latm
P
V = (3.31 mol)
molK (259.2 K) =
4.65 atm
V = 3.31 × .0821 × 259.2 ÷ 4.65 = 15.15 L
The Ideal Gas Law
Ex2: How many grams of propane (C3H8) fit into a 865 mL container
under 528.73 kPa of pressure at 33.7C ?
P = 528.73 kPa _____
5.22 atm
V = 865 mL 0.865
_____ L
n = ? (in g)
306.7 K
T = 33.7 C  ______
PV
nRT
__ = ___
RT RT
528.73 kPa
865 mL
1 atm
= 5.22 atm
101.325 kPa
1L
= 0.865 L
1000 mL
n = (5.22 atm)(0.865 L) = 0.18 mol
C3H8
.0821 Latm (306.7 K)
molK
n = 5.22 × .865 ÷ (.0821 × 306.7) = 0.18
n = PV
RT
0.18 mol C3H8 44.11 g C3H8 = 7.94 g C3H8
1 mol C3H8