# Video: Work Done in an Isothermal Process

An ideal gas expands quasi-statically to twice its original volume. Determine the ratio of the work done in an isothermal expansion to the work done in an isobaric expansion.

07:05

### Video Transcript

An ideal gas expands quasi-statically to twice its original volume. Determine the ratio of the work done in an isothermal expansion to the work done in an isobaric expansion.

Alrighty! So in this question, we’ve been told we’ve got an ideal gas which expands quasi-statically. The gas expands to twice its original volume. We need to determine the ratio of the work done in an isothermal expansion to the work done in isobaric expansion. Now let’s first look at this really weird word here: quasi-statically. A quasi-static process is a thermodynamic process which happens slowly enough for the system to remain in equilibrium whilst the process is occurring.

In this case, we’re looking at the expansion of an ideal gas. If the expansion of the gas is a quasi-static process, then this expansion happens slowly enough the gas still remains in equilibrium whilst this expansion is occurring. In other words, the gas cannot expand so quickly that the pressure is not evenly distributed throughout the gas. And the fact that the gas is quasi-statically expanding makes life a little bit easier for us, because we can treat it as being in equilibrium throughout the process and also before and after the process.

So let’s move on to look at the bit that tells us that the gas is expanding to twice its original volume. Basically, let’s say that the initial volume of the gas before the expansion is 𝑉 one and the volume after the expansion is 𝑉 two. Well, we know that the volume after the expansion is twice as large as the volume before the expansion because the gas has expanded to twice its original volume. So 𝑉 two is equal to two 𝑉 one. We need to find the ratio after work done in an isothermal expansion to the work done in an isobaric expansion.

We’ve called the work done in an isothermal expansion 𝑊 sub 𝑇 and the same thing for the isobaric expansion is 𝑊 sub 𝑃. So let’s clarify why we’ve used the subscripts 𝑇 and 𝑃. An isothermal process is one in which the temperature remains constant, hence it’s known as isothermal. An isobaric process is one in which the pressure remains constant. Hence we’ve used the subscript 𝑇 to represent the work done in an isothermal expansion and the subscript 𝑃 to represent the work done in an isobaric expansion. And we need to determine this ratio, the ratio between the work done in a constant temperature process divided by the work done in a constant pressure process.

The other important thing to note is that we have an ideal gas. Therefore, the ideal gas equation must apply. On the left-hand side we have the pressure 𝑃 multiplied by the volume of the gas 𝑉. And on the right-hand side, we’ve got 𝑛, the number of moles of gas, multiplied by 𝑅, the molar gas constant, multiplied by 𝑇, the temperature of the gas. Finally, we can recall that the work done for a gas is given by the negative integral between 𝑉 one and 𝑉 two of the pressure 𝑃 with respect to the volume 𝑉. So let’s start by working out 𝑊 sub 𝑇. The work done in an isothermal expansion.

𝑊 sub 𝑇, as we can expect, is given by the negative integral between 𝑉 one and 𝑉 two of 𝑃𝑑𝑉. Now because we’ve got an ideal gas, we know that the pressure is given by 𝑃 is equal to 𝑛𝑅𝑇 over 𝑉. So we can substitute this into our expression for 𝑊 sub 𝑇. Now in this expression, 𝑛, the number of moles of gas that we have, stays constant because we’re not changing this. The amount of gas this constant. As well as this 𝑅 is always a constant anyway; it’s the molar gas constant. And 𝑇 is constant because this is an isothermal process. So we can pull out all the constants from the integral, which leaves us with an integral between 𝑉 one and 𝑉 two of one over 𝑉𝑑𝑉 multiplied by these constants and obviously by negative one.

And we can evaluate the integral, which leaves us with the natural logarithm of 𝑉 two divided by 𝑉 one. And this is the expression for 𝑊 sub 𝑇. So let’s move on to finding an expression for 𝑊 sub 𝑃. Once again, we start with the same expression: the negative integral between 𝑉 one and 𝑉 two of 𝑃𝑑𝑉. However, this time we know that the pressure is constant. It’s an isobaric process after all. And as we said earlier, an isobaric process is one in which the pressure is constant. So this time we can pull out the 𝑃 from the integral, which leaves us with just an integral between 𝑉 one and 𝑉 two of 𝑑𝑉. And this whole thing just simplifies to negative 𝑃 multiplied by 𝑉 two minus 𝑉 one.

Now let’s start simplifying each one of these expressions. Let’s go back to 𝑊 sub 𝑇. In the case of 𝑊 sub 𝑇, we’ve got this 𝑛𝑅𝑇 in the expression. Now this 𝑛𝑅𝑇 looks exactly like the 𝑛𝑅𝑇 in the right-hand side of the ideal gas equation. And the whole right-hand side of this equation must be a constant, because as we said earlier the number of moles of gas that we have is constant, 𝑅 is a constant anyway, and the temperature is constant for an isothermal process. So if the whole right-hand side of the equation is constant, then so must the left-hand side be. In other words, the product 𝑃𝑉 must remain constant for an isothermal process.

What this means in practice is that the product of the pressure and the volume before the expansion is equal to the product of the pressure and volume after the expansion. And both of these are equal to 𝑛𝑅𝑇, the constant on the right-hand side of the equation, which means that instead of 𝑛𝑅𝑇 we can choose to substitute either one of these into our equation. Let’s go with 𝑃 one 𝑉 one, because this will make life easier for us later. So we’ve now got an expression that says 𝑊 sub 𝑇 is equal to negative 𝑃 one 𝑉 one multiplied by the natural log of 𝑉 two over 𝑉 one.

The next thing we can do is to remember that 𝑉 one and 𝑉 two are related; two 𝑉 one is equal to 𝑉 two. So we can substitute 𝑉 two in our expression with two 𝑉 one, at which point the 𝑉 ones cancel leaving us with just the natural log of two. And so our final expression for 𝑊 sub 𝑇 is that 𝑊 sub 𝑇 is equal to negative 𝑃 one 𝑉 one multiplied by the natural log of two. We can then move on to 𝑊 sub 𝑃. Once again, we saw earlier that 𝑉 two is equal to two 𝑉 one. So we can substitute that in. And this leaves us with just a singular 𝑉 one. Hence 𝑊 sub 𝑃 is equal to negative 𝑃 multiplied by 𝑉 one. But we can notice that this 𝑃 doesn’t have a subscript, so which 𝑃 is it referring to?

Well in an isobaric process, which is what we’ve got here, we know that 𝑃 is a constant. In other words, the pressure before the expansion is equal to the pressure after the expansion; 𝑃 one is equal to 𝑃 two. And this is what we’re calling 𝑃, the pressure before and after the expansion. So we can choose to replace 𝑃 with either 𝑃 one or 𝑃 two if we want to. And here, it makes sense to replace it with 𝑃 one because we’re going to be dividing 𝑊 sub 𝑇 by 𝑊 sub 𝑃. And so we want things to cancel out.

In fact, let’s do that now! 𝑊 sub 𝑇 divided by 𝑊 sub 𝑃 is equal to negative 𝑃 one 𝑉 one times the natural log of two divided by negative 𝑃 one 𝑉 one. And as we can see all of this cancels out, leaving us simply with the natural log of two. And hence our final answer is that the ratio of the work done in an isothermal expansion to the work done in an isobaric expansion is equal to the natural log of two.