During the isobaric expansion from state 𝐴 to state 𝐵 represented in the diagram, the gas heats its surroundings with 599 joules. What is the change in the gas’s internal energy?
Okay, so in this question, we can see that we’ve got an isobaric expansion from state 𝐴 to state 𝐵. And we can actually see that this is isobaric because it all happens at a constant pressure. And that’s what isobaric means — it happens at a constant pressure. We’re also told that the gas heats its surroundings with 599 joules.
We need to find the change in the gas’s internal energy. To do this, let’s start by recalling the first law of thermodynamics: Δ 𝑢, the change in internal energy of the system, is equal to 𝑄, the heat supplied to the system, plus 𝑊, the work done on the system. Now, it’s very important to note that 𝑄 is the heat supplied to the system and 𝑊 is the work done on the system. Because if the heat’s supplied by the system to its surroundings, then the value of 𝑄 becomes negative. Similarly, if the work is done by the system onto the surroundings, then the value of 𝑊 also becomes negative.
Now, in this question, our system is the gas that’s undergoing isobaric expansion. And we’ve also been told that the gas heats its surroundings. In other words, the gas is giving out heat. So the value of 𝑄 is negative because the heat supplied to the system is negative 599 joules because the heat is being supplied by the system to the surroundings. So we’re halfway to solving our problem already. We already know the value of 𝑄. But we don’t know yet however is the value of 𝑊 — the work done on the system.
To calculate this, we can recall that the work done on the system is given by the integral between the initial and final volumes of the pressure with respect to the volume. In other words, 𝑊 is equal to the integral between 𝑉 one and 𝑉 two of 𝑝𝑑𝑉. So we can work this out. We can say that 𝑊 is equal to the integral between 0.2 meters cubed and 0.45 meters cubed because that’s the initial and final volume, respectively. And the integral is of the pressure, which in this case is 3.0 times 10 to the power of four newtons per meter squared. And of course, this integral is taken with respect to the volume.
Now, the pressure stays constant. So we can actually pull this out of the integral. What that leaves us with is the value of the pressure — the constant pressure outside the integral — and then inside the integral, we’ve just got one. Evaluating the integral — that’s integrating one with respect to the volume — gives us a value of 𝑉. And we have to evaluate this between 0.45 meters cubed and 0.2 meters cubed.
And so the part inside the parenthesis becomes 0.45 meters cubed minus 0.2 meters cubed, which simplifies to 0.25 meters cubed. And hence multiplying that by the constant pressure on the outside of the parenthesis, we get that the work done is 7.5 times 10 to the power of three newton meters.
But we also know that newton meters is the same as the unit of joules which is the unit of energy. Now, we know that the value of 𝑊 that we have here must be positive because the gas is expanding. So we have to do work on the gas — the surroundings have to do the work on the gas — in order to expand the gas — in order for the particles to separate.
And so at this point, we’ve got everything we need to calculate the change in internal energy. We said earlier that Δ 𝑢, the change in internal energy, is equal to 𝑄 plus 𝑊. This ends up being negative 599 joules plus 7.5 times 10 to the power of three joules. Evaluating this gives us a value of 6901 joules. But remember the values that we’ve been given at the question are all either to two significant figures or to three significant figures. So we need to give our answer to the lowest number of significant figures that we’ve been given in the question. And that number is two significant figures. So we need to give our answer to two significant figures.
And so our final answer is that the change in the gas’s internal energy is 6900 joules.