Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 16: 27 February

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Ideal Gas Law, 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

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. Ptot = Pindiv 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

PO2= Ptot - PH2O

PO2= 762 mmHg - 23.8 mmHg (PH2O from Table 10.8, p 486 of Zumdahl)

PO2= 738.2 mmHg, but need atm to use Gas constant,

Patm = (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.

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.

Equation:  C8H18 + O2 right arrow CO2 + H2O
Balancing:  2 C8H18 + 25 O2 right arrow 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

(Do you think you know how to figure out which is limiting? Try estimating. I would approximate 12.5 = 10 and calculate in my head whether have enough oxygen given the octane volume.)

Note the volume change in this reaction: start with 0.625 L end up with 0.7125 L
Equation:  C8H18 + O2 right arrow CO2 + H2O
Balancing:  2 C8H18 + 25 O2 right arrow 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

  • V octane required to react with 0.600 L oxygen = {2/25}{0.600 L} = 0.048 L, more than available, so octane limits;
  • O2 = 0.600 - (25/2)(0.025) = 0.600-0.3125 = 0.288 L,
  • From V octane can then find V CO2 = (16/2)(0.025) = 0.200 L and V H2O = (18/2)(0.025) = 0.225 L
After reaction: 0 L   0.288 L   0.200 L   0.225 L

Now let's 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)

We will finish this next period.

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© R A Paselk

Last modified 2 March 2015