In our exploration to describe the behavior of gases, we come upon the ideal gas law. Now the ideal gas law formula relates the behavior of gas under varied conditions of pressure, volume, moles, and temperature. So our ideal gas law formula is PV=nRT. We see that it's in a purple box, which means that you need to memorize it. Your professor is going to expect you to know this formula. It's going to be involved in so many different types of ideas and calculations within this chapter. Now if we take a closer look at it, we said that P equals pressure. Here, the units for pressure will be in atmospheres. Volume will be in liters, n equals the amount of gas in moles. R is a gas constant, and here when it comes to the ideal gas law, we tend to see it as 0.08206. In the next section, we'll talk about the units involved with this gas constant. And then temperature here, temperature will be in Kelvins. So just remember, to talk about the ideal behavior of gases, we use the ideal gas law. So click on the next video, and let's take a look at an example question.
The Ideal Gas Law - Online Tutor, Practice Problems & Exam Prep
The Ideal Gas Law formula relates the behavior of gases under varying conditions of pressure, volume, moles and temperature.
The Ideal Gas Law
The Ideal Gas Law
Video transcript
The Ideal Gas Law Example 1
Video transcript
Here we're told that a 500 milliliter container at a pressure of 600 milliliters of mercury possesses 29.3 grams of nitrogen gas at 50 degrees Celsius. Here we need to determine the correct units needed for the ideal gas law. So remember, the ideal gas law is PV=nRT. So P stands for pressure. Remember pressure needs to be in units of atmospheres. So here, we're going to do a conversion. We're going to go from 600 millimeters of mercury to atmospheres. Millimeters of mercury go here on the bottom, and then atmosphere goes on top. Remember the relationship is that for every one atmosphere, it's 760 millimeters of mercury. So when we do that, we're going to get a pressure of 0.789 atmospheres. Although 600 has one significant figure, let's just go with three significant figures here, just so we don't have just 0.8 as atmospheres.
Next, volume. Volume needs to be in liters, so we have 500 milliliters. And remember, 1 milli equals \(10^{-3}\). So this would just come out as 0.500 liters.
Next, we need moles, so moles is n. Remember nitrogen gas. Nitrogen in its natural or standard state is N2. It's one of our diatomic molecules. So we're told that we have 29.3 grams of N2. We must convert that into moles, so 1 mole of N2. There are 2 nitrogens. According to the periodic table, each is 14.01 grams. So multiplying that by 2 gives me 28.2 grams for the combined weight in N2. So grams here cancel out and I'll have my moles. Here we'll put it as 1.05 moles of N2.
And then finally, we need our temperature. R is our constant, so we don't have to worry about providing the correct units. It's 0.08206 like we said earlier. So temperature here needs to be in Kelvin. So remember to go from Celsius to Kelvin, you're supposed to add 273.15. Sometimes professors will just say plus 273, but if you want to be as accurate as possible, you want to put 273.15. So when you add that, that's 323.15 Kelvin. So here we have all of the units that we need in the correct forms. So just remember the ideal gas law and what are the correct units that we need to utilize for each one of these variables.
The Ideal Gas Law
Video transcript
As we said, the ideal gas law uses r, which is our gas constant, but realize that the gas constant r can have 2 different values depending on the situation. Now the 2 terms you'll tend to see when it comes to the r constant is the first one we talked about earlier, r being equal to 0.08206. We utilize this value when dealing with the ideal gas law, but r can also equal 8.314. This happens when we're dealing with speed, velocity, or energy. So if a question is dealing with the speed of an object, or a subatomic particle, or dealing with the energy of a reaction or solution, the r converts to 8.314. Now the conversion factor between the r groups is that 1 liter times atmospheres equals 101.325 joules. So we have our 0.08206 here. We use the conversion factor. So remember, we want to get rid of liters times atmospheres, so we put them on the bottom. So 1 liter times atmosphere here on the bottom is 101.325 joules here on top. Liters times atmospheres cancel out and that's how we end up with joules at the end. When you plug this in, you get a long string of numbers as r8.3147295J/mol∙K joules over moles times K. But we just focus on the 8.314 portion. I know there's a 7 there when you work it out completely, but to keep it simple, books will just list it as 8.314. So just keep in mind the gas constant can be 2 different numbers. And remember with the ideal gas law, we use this value, and when dealing with speed, velocity, or energy, we use this value of 8.314.
The Ideal Gas Law Example 2
Video transcript
Here in this example question, it says, how many moles of ammonia are contained in a 25-liter tank at 190 degrees Celsius and 5.20 atmospheres. Alright. So they're asking me to find moles of ammonia. So that would be my n variable. They're giving me liters, so that would be volume. They're giving me degrees Celsius which is temperature, and they're giving me atmospheres which is a unit for pressure. We can see that we have PV = nRT, so we have the ideal gas law to deal with. And the variable that's missing, the one that we're looking for is moles. So all we have to do here is isolate moles. So divide both sides here by RT, and then we're going to say here that moles equals PV over RT. Pressure is already in atmospheres, so no need to convert. Then we're going to have here volume which is already in liters so no need to convert. Then we have our R constant, remember, is 0.08206. And now that we know that we're using the ideal gas law, we have to use the units of liters times atmospheres per moles times Kelvin.
Temperature, remember, has to be in Kelvin. So there is a little bit of converting here necessary. So we have 190 degrees Celsius, so add 273.15 to that and that'll give us 463.15 Kelvin. So take those and plug it in, 463.15 Kelvin. So what cancels out? Atmospheres cancel out, liters cancel out, Kelvins cancel out, and you'll see that the only unit left are the moles that we need to find. So we're going to get initially 3.4205 moles. If we look back on the question, this has 3 significant figures, 2 significant figures, and 3 significant figures. We go with the least number of significant figures, so we're going to put our answer as 3.4 moles of ammonia. So this would be our final answer. So just realize what variables you are being given, what's the missing variable, just use your algebra skills to isolate that one variable, and you'll get your answer at the end.
How many grams of carbon dioxide, CO2, are present in a 0.150 L flask recorded at 525 mmHg and 32 ºC?
1.77 g
0.93 g
0.66 g
0.18 g
0.052 g
How many liters of HNO3 gas, measured at 28.0 ºC and 780 torr, are required to prepare 2.30 L of 4.15 M solution of nitric acid?
When 0.670 g argon is added to a 500 cm3 container with a sample of oxygen gas, the total pressure of the gases is found to be 1.52 atm at a temperature of 340 K. What is the mass of the oxygen gas in the bulb?
