In this article, I describe how easily ideal gas law problems can be solved. Again, I preach the same message: Begin each problem with an appropriate fundamental principle statement. The trick here is very simple: just start the solution of each problem with the ideal gas law expressed in a certain way. This approach is explained in some detail in the first problem solved. Its usefulness is then demonstrated in two additional problems in this article and in two seemingly difficult problems in the next article. In all cases, Boyle’s law, Charles’s law, and the many other forms that the ideal gas law takes in its various applications are not mentioned. Each problem solution is approached in the same way.
Since the text editor doesn’t support certain math symbols, I must use some unusual notation: (a) Almost all variables are represented by an uppercase letter with lowercase letters used as subscripts. For example, the initial pressure (i) (P) is represented by Pi. (B) Powers are designated by ^. For example, “X squared” is written X ^ 2 and “2.2 times 10 raised to the fifth” is written 2.2 x 10 ^ 5.
Problem. A container with a volume of 2.00 L contains 3.00 moles of an ideal gas at a temperature of 293 K and a pressure of 36.1 atm. The gas is compressed to a volume of 1.00 L while its temperature is kept constant by placing the container in a large tub of water at 293 K. What is the new pressure of the gas?
Analysis. With the initial and final state of the gas designated by (i) and (u), we have from the ideal gas law
……………………………….. Ideal Gas Law
………………………… PiVi / NiTi = R = PuVu / NuTu.
In this particular case, a fixed quantity of gas is compressed at a constant temperature; then Ni = Nu and Ti = Tu, and the above equation reduces to
………………………………… PiVi = PuVu.
With Pi = 36.1 atm, Vi = 2.00 L and Vu = 1.00 L,
……………. Pu = PiVi / Vu = (36.1 atm) (2.00 L) / (1.00 L) = 72.2 atm.
This problem shows a simple method that you can use over and over again when applying the ideal gas law. If you are comparing two states (say states (i) and (u)), just write
…………………………….. Ideal Gas Law
…………………… PiVi / NiTi = R = PuVu / NuTu.
Then ask: “What is the same for both states?” Variables that are equal can be canceled on both sides of this equation. For example, in this problem, the number of moles and the temperature are the same. Consequently, those two terms cancel out, leaving
………………………………………. PiVi = PuVu.
Problem. The container in the previous problem has a small leak and the gas is slowly escaping. The leak is discovered and repaired, and the gas pressure is measured at 20.4 atm when the container is in water at 293 K. How much gas is left in the container?
Analysis. This time, Vi = Vu and Ti = You, then
……………………… Ideal Gas Law
………………. PiVi / NiTi = R = PuVu / NuTu
reduce to
……………………….. Pi / Ni = Pu / Nu,
and ………………… Nu = PuNi / Pi = (20.4 atm) (3.00 mol) / (36.1 atm) = 1, 70 mol.
Problem. A car tire is filled to a gauge pressure of 2.10 x 10 ^ 5 N / m ^ 2 when the air temperature is 300 K and the atmospheric pressure is 1.00 x 10 ^ 5 N / m ^ 2 The car is driven on the road and the gauge pressure increases to 2.30 x 10 ^ 5 N / m ^ 2. Assuming that the volume of the tire does not change, what is the temperature of the air inside the tire at the pressure plus high?
Analysis. The absolute pressure inside the tire increases from (2.10 + 1.00) x 10 ^ 5 N / m ^ 2 = 3.10 x 10 ^ 5 N / m ^ 2 to (2.30 + 1.00) x 10 ^ 5 N / m ^ 2 = 3.30 X 10 ^ 5 N / m ^ 2. The volume and number of moles of air do not change when the tire is heated. Using
……………………….. Ideal Gas Law
………………… PiVi / NiTi = R = PuVu / NuTu
with Vi = Vu and Ni = Nu, we have
………………………… Pi / Ti = Pu / Tu.
Due,
……………………. Tu = PuTi / Pi = (3.30 x 10 ^ 5 N / m ^ 2) (300 K) / (3.10 x 10 ^ 5 N / m ^ 2) = 319 K.
Notice how easily these problems were solved. I’m sure that if you can get your students to use the approach I’ve described here, their ability to work with the ideal gas law will be greatly improved.
