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
 P_{tot} = 762 mmHg, but this measured P 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 (P_{H2O} from Table 10.8, p 486 of Zumdahl)
P_{O2}= 738.2 mmHg, but need atm to use Gas constant,
P_{atm} = (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.
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 (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.
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 
(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:  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


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 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})
We will finish this next period.
Syllabus / Schedule 
© R A Paselk
Last modified 2 March 2015