The pressure and volume of a gas is changed in a process, as shown in the accompanying diagram. The curved section between 𝑅 and 𝑆 is semicircular. What work is done by the gas? If the process is reversed, what work is done by the gas?
Looking at this diagram, we see it’s a plot of pressure, in units of atmospheres, versus volume, in units of litres, of a gas. The graph in blue on this plot shows us how these values change. And based on these changes, we want to solve for the work done by the gas. When this process moves in the clockwise direction as shown, we’ll call the work done by the gas 𝑊 sub 𝑐𝑤. And if the process is reversed, meaning we’re moving counterclockwise instead of clockwise, we’ll call the work done by the gas under that condition 𝑊 sub 𝑐𝑐𝑤.
Whether the process moves in a clockwise or counterclockwise direction, regardless, we’ll want to recall that work done by a gas is equal to the area under the curve on a 𝑃𝑉 diagram like we have here. When we consider the clockwise direction, there are two segments we’ll want to consider. The first segment is the semicircular arc that moves from 𝑅 to 𝑆. And the second part is the linear segment that returns back to 𝑅. Based on our diagram, we can solve for the area under the curve for these two segments and therefore calculate the overall work done by this gas.
Considering the area under the first segment of our curve, we see it consists of the area of the semicircle as well as the rectangular area beneath that. Then the area under our second segment is simply that rectangular area again. This tells us that the work done by our gas in a clockwise direction equals the area of the semicircle plus the area of the rectangle then minus the area of that same rectangle. The reason there is a minus sign in front of this second rectangular area is because on the second leg of our journey, we’re moving right to left, counteracting the sign of the work done as we move left to right.
Knowing this then, we can simply cross out this area of the rectangle since it cancels out. 𝑊 sub 𝑐𝑤 then equals simply the area of the semicircle. And recalling that the area of a circle is equal to 𝜋 times its radius squared, we can say the area of the semicircle is one-half that value. When we look at this circle, we see that the radius of the circle is equal to 1.0. But the question is, 1.0 what? What are the units? It seems that if we consider the radius pointing up and down, then our units would be atmospheres. By considering the radius left to right, they might be litres.
It’s helpful to recall that we’re calculating an area. Therefore, the units of our answer must be equal to the units on our vertical axis multiplied by the units on our horizontal axis. Since we’re taking this radius value and then squaring it, that means we can write our radius value as 1.0 with units of square root atmosphere times litres. If we square this result as we plan to square 𝑅, then we’ll end up with units of atmosphere litres, just as we expect for an area.
Plugging in for 𝑅 then, we find a result of 𝜋 over two times 1.0 atmosphere litres or 𝜋 over two atmosphere litres. This is indeed a result for the work done in the clockwise direction. But, since we’re used to seeing work in units of joules, it’ll be nice to convert this answer so we can express it that way.
To do this, we can convert the pressure units from atmospheres into pascals, recalling that one atmosphere is equal to 101325 pascals. And further, we’ll convert the volume units of litres into meters cubed. One litre is equal to 0.001 cubic meters. When we make these unit substitutions, pascals for atmospheres and cubic meters for litres, we see we’ll have a new value to calculate for 𝑊 sub 𝑐𝑤. To two significant figures, it’s 160. And the units are joules. This is how much work the gas does, going through this process in a clockwise direction.
For our next step, as we consider this process, we want to reverse the direction of the process, so it moves now in a counterclockwise direction, and answer the same question: how much work is done by the gas? Changing only the direction of our thermodynamic process means that all the terms we had before are still there. Now, they just have different signs in front of them.
As before, the area of our rectangle cancels out. And we’re left just with the area of the semicircle. But now, it’s minus that area. We calculated that area before. It’s 160 joules. So that means when we reverse direction, we get the opposite value, negative 160 joules. That’s how much work the gas does when the process changes direction.