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?

• 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. 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

• V = 234 mL = 0.234 L
• R = 0.0821 L*atm/mol*K
• T = 25 + 273.15 = 298.15 K
• Ptot = 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.

PO2= Ptot - PH2O

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

PO2= 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:
 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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