chapter 12

GasLaws

Chapter 12Molecular Composition of Gases

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First, Let’s Review Chapter 9-11

The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometric Calculations

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• From the mass of Substance A you can use the ratio of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

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Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

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Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of

first (in this case, the H2)

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Limiting Reactants

• In the example below, the O2 would be the excess reagent

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12.1 Volume - Mass Relationships of Gases

• Last chapter, we related the volume and the mass of a gas:– Reaction Stoichiometry

• Note: a “stoich” problem can be recognized by the fact that information from one substances is given and the problem is asking about a different substance.

– The Combined Gas Law

• Let’s compare the volumes of gases in two example problems…

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…Reaction Stoichiometry at standard conditions

• Example 1 - How many L of Oxygen are needed to react with 50.0L of Hydrogen at STP?

2H2(g) + O2(g) ---> 2 H2O(g)

– But...we can skip steps 1 and 3.– Why? All gases take up the same amount of space at

STP.

50.0 L H2 x1 mole H2

22.4 L H2

x1 mole O2

2 moles H2

x22.4 L O2

1 mole O2

= 25.0 L O2

1 32

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• Example 2 - Calculate the volume of one mole of H2 at 20ºC and 1000 torr.

…The Combined Gas Law at nonstandard conditions

P1 = 760 torr P2 = 1000 torr

T1 = 273K T2 = 293K

V1 = 22.4 L V2 = ?

P1V1 = P2V2

T1 T2

(760)(22.4) = (1000)(V2)

(293) (273)

V2 = 18.3 L H2

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12.2 The Ideal Gas Law• We solved example 2 using the combined gas law – we could

have also used• A relationship between pressure, volume, temperature and the #

of moles of a gas• A new formula that can help us solve all kinds of gas law

problems more easily. PV = nRT

n = # of

molesin Liters

Volume R = ideal gas constant

Kelvin

Temp.

1 atm

1 atm = 760 torr

Pressure

at STP, R = (1atm)(22.4L / 1mol)(273K)

= .0821 L · atm / mol· K

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Deriving the Ideal Gas Law

• Let’s derive the ideal gas law and gas constant...– Volume is proportional to 1/P (as P is reduced, the V increases)

– V is proportional to T (as T increases, the V increases)

– V is proportional to n (as more moles are added, the V increases)

so...V is proportional to 1P x T x n

so... V (k)(T)(n)

P

so... V RTn

P

so... PV nRT

nT

PVR

)273)(1(

)1)(4.22(

Kmole

atmLR

Kmole

atmL

10821.

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In general…

• The combined gas law, P1V1 = P2V2

T1 T2

is used for changing conditions.• But, a new gas law, the Ideal Gas Law, can be

introduced when you have problems containing:– one set of conditions– solving for grams– solving for moles– calculating molecular weight (molar mass)– calculating density– involving stoichiometry and non-STP conditions

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One Set of Conditions

Example: Calculate the volume of 1.00 mole of Hydrogen at 20.0 ˚C and 1000 torr

Step 1: Convert your pressure to atm.

Step 2: Write the ideal gas law and derive Volume.

Step3: Using V = nRT/P, we have

PV = nRT

V = nRT/P

1000 torr x 1 atm / 760 torr = 1.316 atm

V = (1 mole)(.0821 L atm/mol K)(293K)/1.316 atm

V = 18.3 L H2

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Solving for Grams

Example: Calculate the number of grams of helium in a 6.0 liter cylinder at 27˚C and 800 torr.

Step 1: Convert your pressure to atm and your temperature to Kelvin.

Step 2: Write the ideal gas law and derive for moles.

PV = nRT

n = PV/RT

800 torr x 1 atm / 760 torr = 1.05 atm27˚C + 273 = 300K

Step 3: Calculate the moles.

n = (1.05 atm)(6.0 L)/(.0821 L•atm/mol•K)(300K)

n = 0.26 molesStep 4: Convert to grams.

.26 moles x 4.0 g/mole = 1.04 g He

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Solving for Moles

Example: A sample of CO2 in a 10.0 L container at 293K exerts a pressure of 50,000 torr. How many moles of CO2 are in your sample?


Step 2: Write the ideal gas law and derive for moles.

PV = nRT

n = PV/RT

50,000 torr x 1 atm / 760 torr = 65.8 atm

Step 3: Calculate the moles.

n = (65.8 atm)(10.0 L)/(.0821 L•atm/mol•K)(293K)

n = 27.3 moles CO2

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m.w. = g/mol

Calculating M.W.

Example: If 18.0 grams of a gas at 380 torr and 546.0 K occupies 44.8 L, what is the molecular weight of the gas?

Step 2: Identify the m.w. formula. So, plug in 18.0 grams into the formula.

m.w. = 18.0 g/ ? mol

Step 3: Now, use PV = nRT to solve for n (the number of moles).

n = PV/RT

n = (0.50 atm)(44.8 L)/(.0821 L•atm/mol•K)(546 K)

n = .499 moles


380 torr x 1 atm / 760 torr = 0.50 atm

m.w. = 18.0 g/ .499 moles = 36.0 g/moles

Step 4: Plug this value into the molecular weight (m.w.) equation.

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Step 2: If it is assumed that one mole of CO2 is present, then the mass can be calculated from the periodic table and the volume at STP is 22.4 L.

Calculating Density at STP

Example: Find the density of carbon dioxide at STP.

Step 1: Identify the density formula.

D = m/V

D 44.0g

22.4 L

196gL

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D 44.0g

? L

Step 2: Assume one mole of CO2. Thus, CO2 weighs 44.0 grams. Plug this into the Density formula.

Calculating Density (not at STP)

Example: Find the density of carbon dioxide at 546.0 K and 4.00 atm.

Step 1: Identify the density formula.

D = m/V

Step 3: Because the conditions are non-STP values, 22.4 L can’t be used. So, use PV=nRT and solve for V (assume one mole).

V = nRT/P

V = (1.00 mole)(.0821 L•atm/mol•K)(546K)/(4.00 atm)

V =11.206 L

D = 44.0 g/ 11.206 L

Step 4: Plug this value into the density equation.

= 3.93 g/ L

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Deviations from Ideal Behavior

• Real gases do not behave according to the KMT - Why?– Real gases have molecules that occupy space– Real gases have attractive and repulsive forces

• Ideal gases conform exactly to the KMT– no such gas exists– gases only behave close to ideally at low P and high T.– At low T and high P, gases deviate greatly from ideal behavior.

• Some gases are close to ideal (if they are small and nonpolar):– H2, He, Ne (these are small and nonpolar!!!)– O2 and N2 are not too bad– NH3, H2O are not even close to ideal

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12.3 Stoichiometry of Gases

• At STP, there is nothing new here. • The only thing new is that 1 mole = 22.4

L must be adjusted if not at STP.• Problems for Stoichiometry of Gases

include converting: – Grams to Liters– Liters to Grams– Liter to Liter

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Grams to Liters

Example: How many liters of O2 are generated when 50.0 grams of sodium chlorate

decomposes at 0.950 atm and 20.0˚C?

Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole).

Step 2: Plug this value into the appropriate step of the stoichiometry problem.

2NaClO3(s) + heat ---> 2NaCl(s) + 3O2(g)

V = (1 mole)(.0821 L atm/mol K)(293)/.950 atm

V = 25.321 L O2

PV = nRT

V = nRT/P

50.0 g NaClO3 x1 mole NaClO3

106.5 g NaClO3

x3 molesO2

2 moles NaClO3

x25.231L O2

1 moleO2

17.8LO2

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Liter to Grams

Example: If a lawn mower engine generates 555.0 L CO2 on a lovely Sunday

afternoon (.996 atm and 37.00 C0 )- how many grams of octane were consumed?


Step 2: Plug this value into the appropriate step of the stoichiometry problem.

2C8H18(l) + 25O2(g) ---> 16CO2(g) + 18H2O(g)

V = (1 mole)(.0821 L atm/mol K)(310K)/.996 atm

V = 25.551 L CO2

V = nRT/P

2555 COL 1885.309 HCg188

188

2

188

2

2

1

0.114

16

2

551.25

1

HCmole

HCgx

COmoles

HCmolesx

COL

COmole

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Liter to Liter (not at STP)

Example: How many L of carbon dioxide can be made from the combustion of 2.00 L

of propane (C3H8) at 500. K and 3.00 atm? 4C8H3(l) + 18O2(g) ---> 16CO2(g) + 6H2O(g)


V = (1 mole)(.0821 L atm/mol K)(500K)/3.00 atm

V = 13.7 L CO2

V = nRT/P

Step 2 : Plug this value into the appropriate step of the stoichiometry problem.

2.00 L C3H8 x1 mole C3H8

13.7 L C3H8

x16 moles CO2

4 moles C3H8

x13.7 L CO2

1 mole CO2

= 8 L CO2

1 32

– Again...we can skip steps 1 and 3.

chapter 12

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