Unit 3: Topic 4 - Ideal Gas Law Flashcards
Intuitively give an explanation for Boyle’s, Charle’s, and Gay-Lussac’s law
given the ideal gas equation, PV = nRT.
PV = nRT: from this equation, we can break it down into a couple of things.
If n, R, T are constant, then PV is constant, so P and V are inversely related. (Boyle’s Law)
If V, n, R are constant, then P ~ T (~ for proportional), so that P and T are directly related. (Gay-Lussac)
If P, N, R are constant, then V ~ T, so that V and T are directly related. (Charle’s Law)
Consider an ideal gas with a temperature of 50C at 1atm. What is the temperature if the ideal gas’s pressure doubles to 2 atm? What if the pressure halves to 0.5 atm?
It would be nice to just double the temperature, by Gay-Lussac’s law, but we cannot simply double 50C to 100C because we need to convert to Kelvin first. After converting 50C into 323K, we can safely double this value to get 646K or 373C. Similarly, we must halve the temperature in Kelvin to get 161.5K.
Consider a container consisting of hydrogen and oxygen. If the mole ratio between hydrogen and oxygen is 1:1, what is the ratio of their pressures? What if the mass ratio (in grams) is 1:1? How many moles of hydrogen are in the container if there are 34 moles of gas in total (consider both situations)?
Based on the gas law and the partial pressures, we have that the pressure is directly proportional to the number of moles. Therefore, if the mole ratio between hydrogen and oxygen is 1:1, so is their pressure ratio. Here, the ratio for hydrogen to total is 1/2, so there are 34 * 1/2 = 17 moles of hydrogen.
For mass ratio, note that if there is 1 gram of hydrogen per 1 gram of oxygen, then there would be 16 moles of hydrogen per mole of oxygen (because the molar mass of H2 is 2, the molar mass of O2 is 32, a factor of 16 greater), and there would also be a 16:1 ratio between pressures. Here, the ratio of hydrogen to total is 16/17, so there would be 34 * 16/17 = 32 moles of hydrogen.
Let there be a container consisting of nitrogen and oxygen (assume there are no reactions, just pure N2, and O2). If there are equal masses of nitrogen and oxygen, and there are 15 moles of gas in the container, what is the mass of N2 in the container? If the total pressure of the container is 1atm, what is the contribution of nitrogen and oxygen to the total pressure?
Equal masses of nitrogen and oxygen, but the molar mass ratio is 8/7, with oxygen’s molar mass being
greater. Therefore, there would be an 8/7 ratio of moles of nitrogen to oxygen, and there would be 8
moles of nitrogen (and 8 * 14 = 126 grams of nitrogen) and 7 moles of oxygen in the container. Since
moles are directly proportional to pressure, there is also an 8/7 ratio of pressure from nitrogen to oxygen,
and thus the contribution by nitrogen would be 8/15 atm and by oxygen 7/15 atm. Notice that
8/15 + 7/15 = 1, which satisfies the law of partial pressure.
Plot graphs to display Boyle’s, Charle’s, and Gay-Lussac’s Laws.
Plots can be found here. Note simply that direct proportions result in linear graphs, while inverse proportions result in hyperbolic (decreasing) graphs.
