The Ideal Gas Law: Study with Video Lessons, Practice Problems & Examples

The ideal gas law, represented by the equation ${\mathrm{PV}}^{=}\mathrm{nRT}$, relates pressure (P), volume (V), moles (n), and temperature (T) of a gas. The gas constant (R) can be 0.08206 L·atm/(mol·K) for the ideal gas law or 8.314 J/(mol·K) for energy-related calculations. Understanding these constants and their applications is crucial for solving problems involving gas behavior, thermodynamics, and kinetic molecular theory.

The Ideal Gas Law

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, and 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 Example 1

Here we're told that a 500 milliliter container at a pressure of 600 millimeters 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 atmospheres go 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 1 significant figure, let's just go with 3 significant figures here, just so that 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 have to convert that into moles, so 1 mole of N₂. So there are 2 nitrogens. According to the periodic table, each one is 14.01 grams. So multiplying that by 2 gives me 28.02 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 a constant, so we don't have to worry about giving the correct units. It's 0.0821 like we said earlier. So the 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, it'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

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 8.3147295joulesmoles×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

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 by RT, and then we're going to say here that n=PVRT. 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 over 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 them 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 3.42050 moles. If we look back on the question, this has 3 significant figures, 2 significant figures, 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?

a) 1.77 g

b) 0.93 g

c) 0.66 g

d) 0.18 g

e) 0.052 g

• A. 1.77 g
• B. 0.93 g
• C. 0.66 g
• D. 0.18 g
• E. 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?

• A. 56 L
• B. 62 L
• C. 189 L
• D. 230 L
• E. 262 L

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?

• A. 0.266 g
• B. 0.335 g
• C. 0.621 g
• D. 0.715 g
• E. 1.72 g

Here’s what students ask on this topic:

What is the ideal gas law and its formula?

The ideal gas law is a fundamental equation in chemistry that describes the behavior of an ideal gas. The formula is given by:

$PV=nRT$

where $P$ is the pressure in atmospheres, $V$ is the volume in liters, $n$ is the number of moles of gas, $R$ is the gas constant (0.08206 L·atm/(mol·K)), and $T$ is the temperature in Kelvins. This law helps predict how gases will respond to changes in pressure, volume, and temperature.

What are the units for the gas constant R in the ideal gas law?

In the ideal gas law, the gas constant $R$ is typically given as 0.08206 L·atm/(mol·K). This means that the units for $R$ are liters times atmospheres per mole per Kelvin. These units ensure that when you use the ideal gas law formula $PV=nRT$, all the units cancel out appropriately, giving you a consistent and accurate result.

When should you use the gas constant 8.314 instead of 0.08206?

The gas constant $R$ can take on different values depending on the context of the problem. Use 0.08206 L·atm/(mol·K) when dealing with the ideal gas law, which involves pressure, volume, and temperature of gases. However, use 8.314 J/(mol·K) when dealing with problems related to energy, speed, or velocity, such as in thermodynamics or kinetic molecular theory. The conversion factor between these two values is that 1 L·atm equals 101.325 J.

How do you convert temperature to Kelvins for the ideal gas law?

To convert temperature from Celsius to Kelvins, you use the formula:

$T=T{C}_{}+273.15$

where $T$ is the temperature in Kelvins and $T{C}_{}$ is the temperature in Celsius. For example, if the temperature is 25°C, the temperature in Kelvins would be:

$25+273.15=298.15$

Kelvins. This conversion is essential for using the ideal gas law correctly.

What is the significance of the ideal gas law in chemistry?

The ideal gas law is significant in chemistry because it provides a simple yet powerful equation to predict the behavior of gases under various conditions. It combines several gas laws (Boyle's, Charles's, and Avogadro's laws) into one comprehensive formula:

$PV=nRT$

This law helps chemists understand and calculate how gases will respond to changes in pressure, volume, temperature, and the number of moles. It is crucial for various applications, including chemical reactions, industrial processes, and environmental science.