# 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?

• 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

• Ptot = 762 mmHg, but this measured P 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 (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

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

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.

• First we need to write a balanced equation, determine the stoichiometry, and the before and after situations assuming a stoichiometric mixture and theoretical yield:
 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
• Our problem is more complex of course, but the easiest way to approach it is to first solve it assumimg no temperature or pressure change during the reaction - we'll take care of that later:

(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.)

 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 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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