In this article, I describe how easily problems involving the ideal gas law can be solved. Again, I preach the same message: Start every problem with a statement of the appropriate fundamental principle. The trick here is a very simple one --- just start every problem solution with the ideal gas law expressed in a certain way. This approach is explained in some detail in the first solved problem. Its utility is then demonstrated in two additional problems in this article plus in two seemingly difficult problems in the next article. In all cases, there is no mention of Boyle's law, Charles' law, and the many other forms that the ideal gas law takes in its various applications. Every problem solution is approached the same way.
Since the text editor does not support certain mathematical symbols, I must use some unusual notation: (a) Almost every variable is represented by a capital letter with lower case letters used as subscripts. For example, the initial (i) pressure (P) is represented by Pi. (b) Powers are designated by a ^. As examples, "X squared" is written X^2, and "2.2 times 10 to the fifth" is written 2.2 x 10^5.
Problem. A container of volume 2.00 L holds 3.00 mol 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 held constant by placing the container in a large vat 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 amount of gas is compressed at constant temperature; so Ni = Nu and Ti = Tu, and the previous 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.
There is a simple method displayed in this problem that you can use over and over when applying the ideal gas law. If you are comparing two states (say states (i) and (u)), you just write
...................................Ideal Gas Law
........................PiVi/NiTi = R = PuVu/NuTu.
Then you ask: "What is the same for both states?". Those variables that are the same can then be cancelled 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 are cancelled, leaving
..............................................PiVi = PuVu.
Problem. The container of the previous problem springs a small leak, and gas slowly leaks out. The leak is discovered and repaired, and the pressure of the gas is measured to be 20.4 atm when the container is sitting in the water at 293 K. How much gas is left in the container?
Analysis. This time, Vi = Vu and Ti = Tu, so
...........................Ideal Gas Law
...................PiVi/NiTi = R = PuVu/NuTu
reduces 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 highway and the gauge pressure is found to increase to 2.30 x 10^5 N/m^2. Assuming the volume of the tire does not change, what is the temperature of the air inside the tire at the higher pressure?
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 heats up. Using
.............................Ideal Gas Law
.....................PiVi/NiTi = R = PuVu/NuTu
with Vi = Vu and Ni = Nu, we have
..............................Pi/Ti = Pu/Tu.
Consequently,
.........................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 am certain that if you just get your students to use the approach I have outlined here, their ability to work with the ideal gas law will be greatly enhanced.
Dr William Moebs is a retired physics professor, who taught at two Universities: Indiana-Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how he emphasizes fundamental principles by consulting PHYSICS HELP.
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