April 30, 2013 Positive Attitude Peccadillo: a slight offense or fault; a minor sin

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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). notice the unit for R. It’s complicated!

R = .0821 L



atm mol



K

The Ideal Gas Law

• • for the Ideal Gas Law ONLY, all units for P, V, n, and T must match the units in the R value.

R = .0821 L



atm 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 Cl

2 13.8

was compressed at 3534 mmHg of pressure and -



C, what volume would it have? (1 mol Cl 2 = 70.90 g)

The Ideal Gas Law

Ex1: If 234.68 g of Cl

2 -13.8

was compressed at 3534 mmHg of pressure and



C, what volume would it have?

P = 3534 mmHg  V = n = T = ?

234.68 g -13.8

 C   _____ mol ______ K 3534 mmHg 1 atm = 4.65 atm 760 mmHg 234.68 g Cl 2 1 mol Cl 2 70.90 g Cl 2 = 3.31 mol Cl 2 PV = nRT P P V = nRT P .0821 L  V = (3.31 mol) mol 4.65 atm atm  K (259.2 K) = V =

3.31 × .0821 × 259.2 ÷ 4.65

=

15.15 L

The Ideal Gas Law

Ex2: How many grams of propane (C

3 H 8 ) fit into a 865 mL container under 528.73 kPa of pressure at 33.7



C ?

P = 528.73 kPa  V = n = T = 865 mL  ? (in g) 33.7

 C  528.73 kPa 1 atm = 101.325 kPa 5.22 atm 865 mL 1 L = 0.865 L 1000 mL PV = nRT RT RT n = PV RT

n = 5.22 ×

n = (5.22 atm)(0.865 L) = .0821 L  atm (306.7 K) mol  K 0.18 mol C 3 H 8

.865 ÷ (.0821 × 306.7) = 0.18

0.18 mol C 3 H 8 44.11 g C 3 H 8 1 mol C 3 H 8 =

7.94 g C 3 H 8