2026 AP Physics 2 Free Response #1

Ajouté : 2026-05-09

[00:00:00]All right, we're looking at the 2026 version J FRQ number one of the AP Physics 2 exam.

[00:00:07]So version J is the one that is publicly released. So let's go through it. Sample of N moles in a monatomic gas in a large sealed thermally conducting container with a fixed volume. So thermally conducting means heat can flow in and out. A small sphere of mass MS is in the container.

[00:00:22]As the of mass MS is in the container.

[00:00:25]Volume of the sphere is much less than the volume of the container. Gas is initially at state X with pressure P and volume V as shown in the figure.

[00:00:31]The sphere is initially in thermal equilibrium with the gas. Okay, the gas undergoes a heating process until the gas is in state Y with pressure 3P and the sphere is again thermal equilibrium with the gas. Total energy transferred to the sphere during the heating process is QS.

[00:00:44]Now here we're looking at a graph of the number of atoms unit speed as a function of atom speed for the gas. So it's a Boltzmann kind of distribution, right?

[00:00:52]It's talking about like how fast Uh oh actually speed is on the horizontal axis. It's a little bit weird.

[00:00:59]But no, no, no, I think that's right.

[00:01:00]That's the velocity and then density so to speak or number. Sketch a curve that could represent the number of atoms per unit speed. So this is the big thing they want you to recognize is to say like hey, when it's warming up like there's more kinetic energy overall. So the center is got to be a little bit higher. So it's just got to be any distribution where you got the peak going to the right is probably fine.

[00:01:20]Okay, cuz there's not this is qualitative just that the center peak here is to the right saying on average there's more kinetic energy now than there was before. Okay? Derive an expression for the change in temperature of the gas for the heating process of the gas undergoes as the gas changes from state X to state Y. Express your answer in terms of this. Okay, so now we're we're we're going from state X and we said we were at pressure P and V and we're going to Y. So we want the change in the temperature. Okay, so we want to think about all the things. I sometimes if you're not sure what to do, I just kind of write down everything that I know, right? So we know the change in internal energy is Q plus um plus any work done on the gas. It's not really what's happening. We also know that internal energy is 3/2 nRT for ideal monatomic gas.

[00:02:09]Um and that is equal to 3/2 PV.

[00:02:13]Right? And so, um the change in the energy way you want to think about it is going to be the change is going to be 3/2. N doesn't change. R doesn't change.

[00:02:23]These are constants. It's going to be just the change in the temperature.

[00:02:26]Right? Cuz it is it's going to be T final minus T initial.

[00:02:30]Right? 3/2 nRT final or T in state Y minus T in state X. Right? And that's that's kind of what we want to find there. This is what we want to solve for. Right? That difference in that portion there.

[00:02:43]Okay. So, then um we're looking at N and P. We're probably going to use PV equals nRT to get that that change in temperature.

[00:02:51]So, I'm just writing down a lot of information. None of this is that necessary. Not part of my answer. It's just my thought process. It's what we're kind of going through this. So, the temperature is PV over NR.

[00:03:01]And so, when we do TY minus TX, we're going to do the pressure at Y. Well, the pressure at Y they said What did they say? We're at 3P.

[00:03:10]So, it's 3P. And um the volume doesn't change. Okay? Because we're heating it, but we're not changing the volume. So, 3PV over NR.

[00:03:21]And then in in state X, the pressure was just P. PV over NR. And so, this is just going to be 2PV over NR. And that would be the change in the temperature. Okay?

[00:03:31]So, we didn't really need any of this stuff. Right? That was just sort of But, we might use it later on.

[00:03:36]Okay? Because they talked about QS. So, at some point we're probably going to be talking about internal energy being changed and there's no work being done on the gas. So, we're probably going to get to that at some point. But, um let's take a look. Derive an expression for the specific heat of the sphere. Express your answers in terms of n, m, s, p, v, q, s as appropriate. Begin your derivation by writing a fundamental fundamental physics principle or equation from the sheet here.

[00:04:01]So, here we're going to probably use What Where does the Where does the specific heat? It's the mc delta t, right? That's kind of where we kind of come in. And so, that's our That's our heat flow equation. Q is going to be qs.

[00:04:16]Mass of the sphere, cs is what we're looking for in the change in temperature. We found because we're in thermal equilibrium with the gas, so whatever the change in temperature of the gas is is the same temperature change in temperature that we're going to see for the sphere.

[00:04:30]Right? And so, then the cs then we just solve for multiplied by nr divided by 2pvms like that. Okay? So, that's pretty Just make sure nms, pvq physical constants, that's r, v, and q qs. Okay. Well, that looks good.

[00:04:51]Uh insulated container is filled. So, now it's insulated. Filled with ml specific heat cl. The original sphere submerged in liquid. The sphere has a mass ms where ms is less than ml and the specific heat where cs is less than cl.

[00:05:01]The initial temperature of the sphere is greater than the initial temperature of the liquid. So, now we're putting a hot sphere into a liquid, right? And it's going to flow out later the liquid and the sphere reach thermal equilibrium.

[00:05:12]The absolute values of the change in the temperature of liquid and the spheres are delta tl and delta ts. So, Okay. So, it's a different scenario, by the way, right? So, we're taking a container, we have a liquid, and then we're putting in a little sphere in here, right? And so, um is the the delta ts greater than or less than equal to? So, what's the same between them, right? So, what's going to happen is you have this hot thing, it's going to flow the heat's going to flow out of the sphere, so it's going to cool the sphere. Sphere's going to cool down and the liquid's going to heat up, right? And so, the same heat flow is going to happen. This Q equals mc delta T. Right? We want to know how does the change in temperature compare.

[00:05:53]We want to do this beyond algebraic equations, but I always like to start with here. So, we're going to have the same heat flow, right? But, um the masses are very different and the c l is different. Now, what they're telling you is ms and cs.

[00:06:09]Both m and c are smaller for the sphere.

[00:06:12]So, if they're smaller for the sphere and the q is the same cuz I'm the same amount of heat flow, then the change in temperature is going to be greater for the sphere. So, it's going to be delta ts is going to be greater than the temperature of the liquid. It's because um by conservation of energy or whatever you say it's the same amount of heat, same q flows.

[00:06:35]Same heat, if you don't want to use the word, heat flows from the sphere to the liquid.

[00:06:44]Uh because and and um the liquid is the liquid is insulated.

[00:06:53]So, there's no no So, no extra heat added.

[00:07:01]Because the liquid has a greater mass and uh specific heat.

[00:07:15]It is It takes more It takes more energy to heat up.

[00:07:23]Change delta t.

[00:07:25]So, the sphere delta ts is greater than delta tl. Okay? Because it has a greater The liquid has more mass, so it's harder to heat up and has a greater specific heat, also makes it harder to heat up.

[00:07:40]Right? For the same amount of energy you're going to heat up less. So, that's kind of the explanation you can put there.