Video: Calculating Particle Number Density in an Ideal Gas

A vacuum chamber does not produce a true vacuum, but can produce a very low pressure of 1.00 × 10⁻⁷ N/m² at a temperature of 20.0°C. How many molecules per cubic centimetre remain in the vacuum chamber?

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Video Transcript

A vacuum chamber does not produce a true vacuum but can produce a very low pressure of 1.00 times 10 to the power of negative seven newtons per metre squared at a temperature of 20.0 degrees Celsius. How many molecules per cubic centimetre remain in the vacuum chamber?

Okay, so in this vacuum chamber, we’ve been told that it can produce a pressure of 1.00 times 10 to the power of negative seven newtons per metre squared. We’ll call this pressure 𝑃. Now this pressure is produced at a temperature of 20.0 degrees Celsius, and we’ll call this 𝑇.

What we’ve been asked to do is to find out the number of molecules per cubic centimetre. In other words, how many molecules are there in a volume of one centimetre cubed? And we’ll call this volume 𝑉. As well as this, we’ll call the number of molecules, which is what we’re trying to find out, capital 𝑁. And we’ll put a question mark next to it because we don’t know what it is yet.

Now to find the value of 𝑁, we need to use an equation called the ideal gas equation, because we can assume that the gas that remains in the vacuum chamber is an ideal gas. The main reason that we can do this in this instance is because the pressure is really, really low.

What this means is that there are very few molecules per every unit volume. Therefore, the space between molecules is actually quite significant compared to the size of these molecules. And so we can say that the average spacing between molecules is quite large.

Now if this average spacing is quite large, then we can assume that each of these particles is basically a point particle. And hence, we can collectively assume of the gas that it’s an ideal gas. And therefore, we can use the ideal gas equation.

Now in the ideal gas equation, the 𝑃 stands for pressure, 𝑉 for the volume occupied, 𝑁 for the number of molecules, 𝑘 is the Boltzmann constant, and 𝑇 is the temperature. We can also recall that the Boltzmann constant is 1.38 times 10 to the power of negative 23 metres squared kilogram per second squared per Kelvin, at which point we know the value of 𝑃, we know the value of 𝑉, we know the value of 𝑘, and the value of 𝑇. And we need to find out what 𝑁 is.

So to do this, we can rearrange the equation. But before we do that, let’s convert all of the quantities that we have to standard units, starting with the pressure. Now the pressure is already in its standard unit of newtons per metre squared. This is because a newton per metre squared is equivalent to a pascal, and that is the standard unit of pressure.

So let’s move on to looking at the temperature. Now this is given in degrees Celsius, but we need to convert it to Kelvin. To do this, we’ll recall that we convert the degree Celsius temperature into Kelvin by adding 273. And so the temperature now becomes 293.0 Kelvin, at which point we don’t need to consider the temperature in degrees Celsius anymore.

So let’s move on to looking at the volume, which has been given to us in centimetres cubed, because we want to work out how many molecules there are per cubic centimetre. Now to do this conversion, we can recall that one metre contains 100 centimetres. And therefore, dividing both sides of the equation by 100, we find that one hundredth of a metre is equal to one centimetre.

So to find out how much one centimetre cubed is in metres cubed, we simply cube both sides of the equation. And this gives us one centimetre cubed as being equivalent to one times 10 to the power of negative six metres cubed, at which point we’ve converted the volume as well into its standard unit of metres cubed.

This means that we can now rearrange the ideal gas equation to solve for 𝑁. Dividing both sides of the equation by 𝑘𝑇, we find that 𝑁 is equal to 𝑃𝑉 divided by 𝑘𝑇. Now we can substitute in the values of 𝑃, 𝑉, 𝑘, and 𝑇, which look something like this.

Now before we evaluate the fraction, let’s discuss the units first, starting with the unit for pressure, which is newtons per metre squared. Now let’s recall that one newton is equivalent to one kilogram metre per second squared because this is Newton’s second law of motion. A unit of force is equal to a unit of mass times a unit of acceleration.

So let’s replace the newton in our fraction with kilogram metre per second squared. This way, this unit of kilogram gets cancelled with this unit of kilogram in the denominator, and this unit of second squared gets cancelled with this unit of per second squared as well. As well as this, we have one, two, three, four powers of metres in the numerator and one, two, three, four powers of metres in the denominator. So those all cancel out. And finally, this unit of per Kelvin cancels with this unit of Kelvin.

So overall, this fraction has no units left then. Well, this makes sense because we’re trying to find the number of molecules, and a number is not going to have a unit. So it seems like we’re on the right track then. This means that we can evaluate the fraction now.

Doing this gives us a value for the number of molecules as 2.47 times 10 to the power of seven molecules per centimetre cubed. And remember, we need to give our answer to three significant figures because the values that we’ve been given at the question, both 1.00 times 10 to the power of negative seven newtons per metre squared and 20.0 degrees Celsius, have been given to three significant figures. So we give our answer to three significant figures as well, at which point we can say we found our final answer. The number of molecules per cubic centimetre that remain in the vacuum chamber is 2.47 times 10 to the power of seven.

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