Chem 109 - General Chemistry - Spring 2011

Lecture Notes 15: 21 February

Gas Laws, cont.

Example: In an experiment the pressure of nitrogen inside a sealed, rigid container is found to be 5.133 atm at 25 °C. How many moles of gas are in the container if it has a volume of 500.0 mL?

Again, start with PV = nRT

P = 5.133 atm

V = 500.0 mL/1000 mL/L = 0.5000 L

n = ?

R = 0.0821 L*atm/mol*K

T = 25° + 273.15° = 298.15 K

Rearranging, n = PV/RT

n = (5.133 atm)(0.5000 L)/(0.0821 L*atm/mol*K)(298.15 K)

n = 1.0485 x 10^-1

n = 0.105 mol

Dalton's Law of Partial Pressures

The pressure of a gas is independent of the presence of other gases. P_tot = ∑ P_{indiv gas}. Seen in Lab. An example is provided for your entertainment.

We should also note that the vapor pressure of a substance in equilibrium with its solid or liquid phase depends only on the substance and its temperature.

Example: In a cyanobacterium photosynthesis experiment, 234 mL of oxygen was collected over water at 762 mmHg and 25°C. How many moles of oxygen was collected?

First we know PV = nRT

Rearranging, n = PV/RT

V = 234 mL = 0.234 L

R = 0.0821 L*atm/mol*K

T = 25 + 273.15 = 298.15 K

P_tot = 762 mmHg, but this is the total pressure due to oxygen and water vapor. We need the pressure due to just the O₂ if we are to find the moles O₂.

P_O₂= P_tot - P_H₂O

P_O₂= 762 mmHg - 23.8 mmHg (from Table 10.8, p 486 of Zumdahl)

P_O₂= 738.2 mmHg

P = (738.2 mmHg) / (760 mmHg/atm) = 0.97132 atm

Now we can substitute and solve for n:

n = (0.97132 atm)(0.234 L) / (0.0821 L*atm/mol*K)(298.15 K) = 9.29 x 10^-3 moles.

Gas Stoichiometry

In chemical equations involving gases we can use volumes in place of molar quantities, since volumes are proportional to moles by Avogadro's Law: V = a*n where a is a constant.


Equation:	C₈H₁₈	+	O₂		CO₂	+	H₂O
Balancing:	2 C₈H₁₈	+	25 O₂		16 CO₂	+	18 H₂O
Stoichiometry (n or V):	2	:	25	:	16	:	18
Before reaction:	1 L		12.5 L		0		0
After reaction:					8 L		9 L

Example: Assume you have a 2.5 L four cylinder engine where the cylinder volume is reduced by a factor of ten in the compression stroke. If 0.025 L of octane (C₈H₁₈) vapor and 0.600 L of oxygen (both measured at 765 mmHg and 25 °C) are introduced into the cylinder, what will be the pressure "at the top of the stroke" if ignition gives a temperature of 557 °C?

We can begin by determining the stoichiometry and thus the amounts of reactants and products left after reaction.

First we need to write a balanced equation, determine the stoichiometry, and the before and after situations.

The easiest way to do this is to assume no temperature or pressure change during the reaction - we'll take care of that later:

Note the volume change in this reaction: start with 0.625 L end up with 0.7125 L
Equation: C₈H₁₈ + O₂ CO₂ + H₂O

Balancing: 2 C₈H₁₈ + 25 O₂ 16 CO₂ + 18 H₂O

Stoichiometry (n or V): 2 : 25 : 16 : 18

Before reaction: 0.025 L 0.600 L 0 0

From the stoichiometry can see that octane is limiting - some octane will be left over. (vol octane required to react with 0.600 L oxygen = {2/25}{0.600 L} = 0.048 L); O₂ = 0.600 - (25/2)(0.025) = 0.600-0.3125 = 0.288 L, CO₂ = (16/2)(0.025) = 0.200 L, H₂O = (18/25)(0.025) = 0.225 L

After reaction: 0 L 0.288 L 0.200 L 0.225 L

Now lets find the pressure as requested:

Use Gas Laws to solve: PV = nRT,

putting constants together, PV/nT = R, or P₁V₁/n₁T₁ = P₂V₂/n₂T₂

Rearranging: P₂ = (P₁)(V₁/V₂)(n₂/n₁)(T₂/T₁)

P₁ = 765 mmHg

V₁ = 0.625 L
V₂ = 0.0625 L

"n₁" = 0.625 L
"n₂" = 0.7125 L

T₁ = 25° + 273 ° = 298 K
T₂ = 557 ° + 273 ° = 830 K

Substituting: P₂ = (765 mmHg)(0.625 L/0.0625 L)(0.7125/0.625)(830 K/298 K) = 2.429 x 10⁴

P₂ = 2.43 x 10⁴ mmHg = 32.0 atm

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© R A Paselk

Last modified 21 February 2011

Humboldt State University ® Department of Chemistry

Richard A. Paselk