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
n = 0.105 mol ^{}
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.
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 O_{2} if we are to find the moles O_{2}.
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.
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: | C_{8}H_{18} | + | O_{2} | CO_{2} | + | H_{2}O | |
Balancing: | 2 C_{8}H_{18} | + | 25 O_{2} | 16 CO_{2} | + | 18 H_{2}O | |
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 (C_{8}H_{18}) 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: C_{8}H_{18} + O_{2} CO_{2} + H_{2}O Balancing: 2 C_{8}H_{18} + 25 O_{2} 16 CO_{2} + 18 H_{2}O 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); O_{2} = 0.600 - (25/2)(0.025) = 0.600-0.3125 = 0.288 L, CO_{2} = (16/2)(0.025) = 0.200 L, H_{2}O = (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 P_{1}V_{1}/n_{1}T_{1} = P_{2}V_{2}/n_{2}T_{2}
Rearranging: P_{2} = (P_{1})(V_{1}/V_{2})(n_{2}/n_{1})(T_{2}/T_{1})
P_{1} = 765 mmHg V_{1} = 0.625 L
V_{2} = 0.0625 L "n_{1}" = 0.625 L
"n_{2}" = 0.7125 L T_{1} = 25° + 273 ° = 298 K
T_{2} = 557 ° + 273 ° = 830 K
Substituting: P_{2} = (765 mmHg)(0.625 L/0.0625 L)(0.7125/0.625)(830 K/298 K) = 2.429 x 10^{4}
P_{2} = 2.43 x 10^{4} mmHg = 32.0 atm
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© R A Paselk
Last modified 21 February 2011