Lesson Video: The Ideal Gas Law in Terms of Number of Moles | Nagwa Lesson Video: The Ideal Gas Law in Terms of Number of Moles | Nagwa

Lesson Video: The Ideal Gas Law in Terms of Number of Moles Physics

In this video, we will learn how to calculate the relationship between the number of moles in an ideal gas and the values of its bulk properties.

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Video Transcript

In this video, we will learn how to calculate the relationship between the number of moles of an ideal gas and the values of its bulk properties. Of course, we’ll be working with a different sort of mole than we see here.

Starting off, let’s note that when we’re working with an ideal gas, that means the gas consists of a large number of identical molecules of negligible size. Each gas particle individually has properties like mass and velocity. But when we’re working with an ideal gas, which as we said is made up of many molecules, often we’re more interested in what are called bulk properties of the gas. These are the properties of pressure, volume, and temperature that apply to the entire gas. Describing an ideal gas in terms of its bulk properties lets us summarize the actions of very many individual molecules.

We know, for example, that gas pressure is due to collisions between individual molecules of a gas in the walls of that gas’s container. Rather than having to account for every single individual collision, we can simply describe the bulk property of the gas’s pressure. If we represent an ideal gas’s pressure using 𝑃, its volume using 𝑉, and its temperature using 𝑇, then these three properties are related through this expression. The pressure of an ideal gas multiplied by its volume is proportional to the gas’s temperature.

To see a bit about how this relationship works, let’s imagine we keep our gas’s container as it is, in other words, we hold the volume of the gas constant, but that we raise the gas’s temperature. Gas temperature, we can recall, is related to the average speed of the gas molecules. An increase in temperature then means the average speed of these molecules will also increase. And therefore, as these molecules of gas collide with the walls of the container, they will be doing so more frequently and at a higher average speed and therefore overall will exert a greater pressure on the container walls.

So then, for an ideal gas, the gas’s pressure multiplied by its volume is directly proportional to the gas’s temperature. Mathematically, whenever we have one quantity being directly proportional to another, we can relate those two quantities using an equation. Specifically, we write that one quantity, in this case 𝑃 times 𝑉, as being equal to a constant — we’ve called that constant 𝑘 — multiplied by the other quantity, in this case temperature 𝑇. 𝑘 is a constant in the sense that it does not depend on gas pressure, volume, or temperature. It does, however, depend on the number of particles in the ideal gas.

At this point, we can recall a common way of describing a number of objects where that number is very large. To do this, we make use of a quantity called the mole. One mole of some kind of object is equal to about six times 10 to the 23 of that object. The quantity of the mole is useful for taking large numbers of objects and making those numbers more manageable. We bring all this up because our constant 𝑘, which we said it depends on the number of gas particles, can be expressed as the number of moles of the ideal gas involved, 𝑛, multiplied by a value, 𝑅, that is called the molar gas constant.

Using this fact that 𝑘 can be written as 𝑛 times 𝑅, we can write our equation describing an ideal gas like this. The pressure 𝑃 of an ideal gas multiplied by its volume 𝑉 is equal to the number of moles of the gas 𝑛 times this molar gas constant 𝑅 multiplied by the gas’s temperature 𝑇. This expression is sometimes called the ideal gas law. Specifically though, it’s the ideal gas law in terms of the number of moles of the gas. Soon enough, we’ll get some practice using this relationship.

For the moment, let’s note that our molar gas constant, capital 𝑅, has an approximate value of 8.31 joules per kelvin mole. The units of 𝑅 may seem strange, but let’s recall that 𝑛 is expressed in units of moles and the temperature 𝑇 is expressed in kelvin. Therefore, when 𝑛 in moles is multiplied by 𝑅 in joules per kelvin mole and that’s multiplied by 𝑇 in kelvin, then the units of moles cancel from numerator and denominator, as do the units of kelvin. We’re left then with units of joules.

A joule, we recall, is equal to a newton times a meter. And because a newton can be expressed as a kilogram-meter per second squared, we can therefore express a joule as a kilogram-meter squared per second squared. All of this means that we have another set of units in terms of which to write our molar gas constant 𝑅. We can write that constant as 8.31 kilograms-meter squared per second squared kelvin mole. This is another way of writing out the same thing as above.

Knowing all this about the ideal gas law in terms of number of moles, let’s look now at an example.

A container of volume 0.225 cubic meters holds 2.24 moles of oxygen gas at a temperature of 320 kelvin. Find the pressure on the container’s interior surfaces. Use a value of 8.31 meter squared kilograms per second squared kelvin mole for the value of the molar gas constant. Give your answer to one decimal place.

Let’s say that this is our container holding oxygen gas. And as we’re told, the volume — we’ll call it 𝑉 — of the container is 0.225 cubic meters and the number of moles of our gas — we’ll call it 𝑛 — is 2.24. Additionally, the gas exists at a temperature 𝑇 of 320 kelvin. Having a temperature above absolute zero, like this temperature is, implies that the oxygen particles in the volume are moving around. As they move, these particles collide both with one another and the walls of the container. The overall effect of these collisions with the container’s interior surfaces is to create a pressure. It’s that pressure that we want to solve for.

And to do it, we’re going to use a form of the ideal gas law, where this form includes the number of moles 𝑛 of the gas involved. This equation says that the pressure 𝑃 of an ideal gas multiplied by that gas’s volume is equal to the number of moles of the gas times the molar gas constant 𝑅 multiplied by the gas’s temperature 𝑇. With our oxygen gas in the container, which we’re treating as an ideal gas, we want to solve for its pressure.

Therefore, we’d like to rearrange this ideal gas law equation so that pressure 𝑃 is the subject. We can do this by dividing both sides of the equation by the gas volume 𝑉. On the left-hand side, the volume 𝑉 in numerator and denominator cancel one another out. And we find that pressure 𝑃 equals 𝑛 times 𝑅 times 𝑇 divided by 𝑉. Our problem statement gives us values for the volume 𝑉, the number of moles 𝑛, and the temperature of the gas 𝑇. Along with this, we’re told a specific value to use for the gas constant 𝑅. We can therefore substitute all four of these values in to our equation for 𝑃.

Let’s note that in our numerator the units of moles cancel out as do the units of kelvin. The remaining units in this expression will be equivalent to the pressure unit of pascals. When we compute this value, we get an answer of approximately 26473 pascals. Let’s recall at this point that 1000 pascals is equal to one kilopascal. If we wanna convert our answer into units of kilopascals then, we take the decimal point and we’ll move it one, two, three spots to the left. We have a pressure then of 26.473 kilopascals. Our question asks us to give our answer to one decimal place. Since the digit two places after the decimal point is greater than or equal to five, that means we’ll round up so that our final answer is 26.5 kilopascals. This is the pressure created on the container’s interior surfaces due to the oxygen gas.

Let’s look now at another example.

A cloud of gas has a pressure of 220 kilopascals and a temperature of 440 kelvin. The gas contains 8.2 moles of a particle with a molar mass of 10.5 grams per mole. Find the volume of the cloud. Use 8.31 meter squared kilograms per second squared kelvin mole for the value of the molar gas constant. Give your answer to two decimal places.

Let’s say that this is our cloud of gas. We’re told the pressure of this gas as well as its temperature. And we know that this gas contains 8.2 moles of the particle that makes up the gas. Knowing all this, we want to solve for the volume, we’ll call it 𝑉, of the cloud. We can begin doing this by recalling the ideal gas law in terms of the number of moles 𝑛 of a gas.

This equation tells us that the pressure of an ideal gas multiplied by its volume equals the number of moles of that gas times the molar gas constant multiplied by the gas’s temperature. As we’ve seen in this example, we want to solve for the gas volume. By dividing both sides of our equation by the gas pressure 𝑃, we cancel that factor on the left. And we see that the gas volume 𝑉 equals 𝑛 times 𝑅 times 𝑇 divided by 𝑃.

We’re going to assume our cloud of gas is an ideal gas and therefore is described by this equation. We can now take the step of substituting in for the four values on the right-hand side of this equation. 𝑛 is equal to 8.2 moles. The molar gas constant 𝑅 is 8.31 meter squared kilograms per second squared kelvin mole, the gas temperature is 440 kelvin, and the gas’s pressure is 220 kilopascals. We’re nearly ready to calculate 𝑉. But before we do, we want to convert the units of our pressure from kilopascals to pascals. We do this so that all of the units in this entire expression are on the same basis.

One kilopascal we recall is equal to 1000 pascals. So if we take this number and multiply it by 1000, then the resulting number 220000 will be our pressure in units of pascals. Now, when we calculate this fraction, we’ll get a result in units of cubic meters. Computing this expression, we find a result of 0.136284 cubic meters. Note though that we’re to give our answer rounded to two decimal places. To that level of precision, our volume is 0.14 cubic meters. This is the volume of our cloud of gas. And note that we didn’t need to use the molar mass of the gas to work this out. Knowing how many moles of gas there are in the cloud was enough.

Let’s look now at one last example.

A gas consisting of 25.6 moles of carbon fills a volume of 0.128 cubic meters and has a pressure of 135 kilopascals. Find the temperature of the gas. Use a value of 12.0107 grams per mole for the molar mass of carbon and 8.31 meter squared kilograms per second squared kelvin moles for the value of the molar gas constant. Give your answer to the nearest kelvin.

Let’s say that this is our gas inside the given volume. Knowing this volume as well as the gas’s pressure and how many moles of gas there are, we want to solve for the gas’s temperature; we’ll call that 𝑇. To begin doing that, we can recall the ideal gas law. This law, written here in terms of the number of moles of our ideal gas, says that if we multiply an ideal gas’s pressure and its volume, then that product is equal to the number of moles of the gas times what’s called the molar gas constant multiplied by the gas’s temperature.

It’s the temperature of our gas that we want to solve for. So, we’ll rearrange this equation to make 𝑇 the subject. Dividing both sides of the equation by 𝑛 times 𝑅, those two factors both cancel out on the right. We see then the temperature of an ideal gas equals its pressure times its volume divided by the number of moles of the gas times the molar gas constant.

In our problem statement, we’re given values for all of the factors that appear on the right-hand side of this equation. The gas pressure 𝑃 is 135 kilopascals; its volume 𝑉 is 0.128 cubic meters. There are 25.6 moles of the gas 𝑛. And the molar gas constant 𝑅 is 8.31 meter squared kilograms per second squared kelvin mole. Before we calculate 𝑇, we want to change the units of our pressure from kilopascals into pascals. We’ll make this change so that all of the units in this expression are on the same footing. That is, they’ll be expressed in units that either are or can be directly converted into SI base units.

One kilopascal we know is equal to 1000 pascals. Therefore, to convert from kilopascals to pascals, we’ll multiply 135 by 1000. We find that in units of pascals, our pressure is 135000 pascals. We’re now ready to calculate the temperature 𝑇. If we round our answer to the nearest whole number, that is, the nearest kelvin, we get a result of 81 kelvin. This is the temperature of the gas.

Note that to find this temperature, we didn’t need to use the molar mass of this gas. Rather, we were able to solve for the temperature using other given information.

Let’s now summarize this lesson through a few key points. In this video, we learned an equation for the ideal gas law in terms of the number of moles of the gas. This equation tells us that the pressure of an ideal gas multiplied by its volume is equal to the number of moles of the gas times what’s called the molar gas constant multiplied by the gas temperature.

The molar gas constant 𝑅 is approximately equal to 8.31 joules per kelvin mole. Written in SI base units, this is equal to 8.31 meter squared kilograms per second squared kelvin mole. Lastly, through a set of examples, we saw that the ideal gas law allows us to solve for bulk properties of a gas, the gas’s pressure, volume, and temperature.

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