Humboldt State University ® Department of Chemistry

Richard A. Paselk
Chem 107

Fundamentals of Chemistry

Fall 2009
Lecture Notes: Gas Law Examples

© R. Paselk 2005
  
 

Worked Examples
  

 Use Back Button to Return to Notes

Example: A 1 L sample of gas weighed 1.25 g at a temperature of 0 °C and a pressure of 1 atm. What is the MW of this gas?

The key here is to first find the number of moles in 1 L since we already know how many grams are in 1 L, so:

PV = nRT, and can re-arrange to

n/V = P/RT = moles/vol

n/(1 L) = (1 atm)/(0.0821 (L-atm)/mole-K))(0 + 273)K

n = 0.04462 mole/L

So we have 0.04462 moles in a 1 L sample which weighs 1.25 g.

Finally, MW = g/mole = 1.25 g/ 0.04462 mole

MW = 28.02 g/mole = 28.0 g/mole

 

 

 

 

 

Example: A student ignores the warning labels and throws an empty (no liquid left, no spray) can of hair spray into his campfire. Assuming an ambient temperature of 25 °C and atmospheric pressure of 7.20 x 102mmHg, and a temperature in the coals of 600 °C, find the pressure in the can in the fire, assuming it doesn't burst or expand.

PV = nRT

First look at the problem to see what changes and what doesn't.

Rearranging to put the variables on one side:

P/T = nR/V, since nR/V is constant, P1/T1 = P2/T2

Rearranging: P2 = (P1)(T2/T1)

Next, convert temperatures into Kelvins

P2 = (720 mmHg)(873 K)/(298 K)

= 2109 mmHg, or in atm

= (2109 mmHg)/(760 mmHg/atm) = 2.77 atm.

 

 

 

 

 

 

 

 

 

 

 

One of the student's colleagues on this ill fated trip tossed an "empty" 0.500 L propane cylinder into the fire. Unfortunately, 3.50 g of propane remained in the cylinder. What pressure would be reached in the cylinder assuming no deformation and no bursting at 550 °C (assume 3 sig figs for the temp.).

PV = nRT

Rearranging, P = nRT/V

P = (7.938 x 10-2 mol)(0.0821 L*atm/mol*K)(823.15 K) / (0.500 L)

P = 10.73 atm

P = 10.7 atm

 

 

 

 

 

Example: 2.40 L of ethene gas (C2H4) is combined with 7.35 L of oxygen and ignited. If all volumes of reactants and products are measured at the same temperature and pressure (above 100 °C - so water is a vapor), calculate the volume of each substance after the reaction is complete.

Equation:  C2H4 + O2 CO2 + H2O
Balancing:  C2H4 + 3 O2 2 CO2 + 2 H2O
Stoichiometry (n or V): 1 : 3 : 2 : 2
Before reaction:  2.40 L   7.35 L   0   0
From the stoichiometry need 3  volumes of oxygen for every volume of ethene, or 3*2.4 = 7.2 L. We have 7.35, so 7.35 - 7.2 = 0.15 L is left over. Each of the products has a 2:1 ratio, so each will be 4.8 L
After reaction: 0 L   0.15 L   4.8 L   4.8 L

 

 

 

 

 

Example: 50.0mL of oxygen is collected over water from a specimen of Anacris water weed illuminated by controlled lighting. If the temperature is 20.0°C and the pressure of the collected gas is 760.5 mmHg how many moles of oxygen were collected? (the vapor pressure of water = 17.5 mmHg at 20.0°C).

PV = nRT

Rearranging, n = PV/RT

First problem is we have to find the pressure of the "dry" oxygen, that is the oxygen gas itself. Since we collected it over water, it will be saturated with water vapor. Thus we need to subtract the vapor pressure of water at 20.0°C given above:

 

C107 Laboratory

C107 Home

C107 Lecture Notes

© R A Paselk

Last modified 18 November 2008