Partial Pressures and Working through a Word Problem

To begin class, Mr. H asked about the webassign that would be due on Friday and whether or not we had any questions on it. The main question was on how to recognize what equations we should use on each problem because, unlike stoichiometry, all our gas equations are word problems that have about 5 key equations. These include:
P1V1=P2V2,
P1/n1=P2/n2,
P1/T1=P2/T2,
PV=nRT (R as a constant),
P1V1/T1=P2V2/T2.
All others can be derived from these core equations and concepts. It is also important to know that if P is divided by a variable, like temperature, then it varies directly with that variable (P/T). If P is multiplied by a variable, like volume, then it varies indirectly with that variable (PV). (look at pgs. 23-24)

After doing a few gas word problems on the white board, we continued on to pg.20. On pg. 20 was a question that seemed very difficult because it incorporated the mixing of two gases and their resulting pressure. We know from the start that the volume of H2 gas is 2L with a pressure of 1.2atm and the volume of the N2 gas is 1L with a pressure of 4.6atm. Temperature and moles stay constant. Although this equations looks very difficult, it is really just a P1V1=P2V2 equation. We begin by solving for 1 gas, H2. First, rearrange the equation using a bit of algebra and make it so that you are solving for P2, our unknown variable. Now, you should end up with P2=P1V1/V2. Next, simply plug in the information we are given to get P2= (1.2atm)(2L)/V2. Because we know that the new volume will be the old volumes added together, V2 is 3L. Now we solve and get that P2=.80atms. Repeat the same process for the N2 gas and you should end up with 1.53atms. Finally, to solve the last problem of the resulting pressure, just add together the two P2 values to get 2.33 atms. The equations should have a diagram that looked something like this:

This is the diagram from the webassign that was due on friday, so no you will not get a free answer!

Finally, we looked at pg. 19, which had similar partial pressure problems (although they are a bit easier). The key to these equations was to use your textbook to look up the vapor pressure of water at certain temperatures. This was given to us by Mr. H. He told us that the V.P. of H20 for problems 1 and 2 where 31.8 and 149 mmHg respectively. This left us with 868.3 mmHg and 151.02 mmHg for the O2 values of questions 1 and 2. Then, to solve parts 3+4 and Q 1-2, we put each pressure value over the total pressure. After converting each kPa pressure to mmHg, the correct answers for Parts 3-4 were 3.53% and 96.46% for Q1 and 49.66% and 50.34% for Q2. The rest of the problems could be solved the same way.

That was all for Thursday's class. Thank You and see you in class!