Consider a well-insulated horizontal rigid cylinder that is...

Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains $1 \mathrm{~m}^{3}$ of $\mathrm{N}_{2}$ gas at $500 \mathrm{kPa}$ and $80^{\circ} \mathrm{C}$ while the other side contains $1 \mathrm{~m}^{3}$ of He gas at $500 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine ( $a$ ) the final equilibrium temperature in the cylinder and $(b)$ the wasted work potential during this process. What would your answer be if the piston were not free to move? Take $T_{0}=25^{\circ} \mathrm{C}$.

Key Concepts

Energy Conservation in Adiabatic Systems

This concept refers to the application of the first law of thermodynamics in a system where no heat or work is exchanged with the surroundings. For an insulated system, the total internal energy remains constant during any process. In problems involving different compartments within an insulated container, the energy lost by one part must be gained by the other, dictating the final equilibrium state.

Ideal Gas Behavior with Constant Specific Heats

This concept involves modeling gases as ideal and assuming that their specific heats (at constant volume or pressure) do not vary significantly with temperature. This assumption simplifies the calculation of changes in internal energy based on temperature differences, allowing for straightforward energy balance calculations in processes where temperature equilibration is observed.

Thermal Equilibrium

Thermal equilibrium is reached when two or more bodies in thermal contact exchange energy until they attain a common temperature. In the context of the segregated cylinder, heat transfer through the piston leads both gases to a final uniform temperature. Energy conservation is used to determine this equilibrium temperature through the averaging of internal energies weighted by the mass or number of molecules in each compartment.

Exergy Destruction and Reversible/Irreversible Processes

Exergy represents the maximum useful work that can be extracted from a system as it comes to equilibrium with its surroundings. In any irreversible process, such as the spontaneous heat transfer between gases, some of this work potential is lost as wasted work. Analyzing exergy destruction helps quantify the inefficiencies and the lost opportunities for performing work, which depends on whether the piston is free to move or held fixed, affecting the process path and available work.

Transcript

00:01 In this problem, we have a tank, and it has a piston in the middle of it.

00:09 And the piston can slide.

00:11 Well, in one case, it can slide, and in the other case, it can't.

00:13 It's fixed.

00:15 But we also have heat transfer across there.

00:19 So in this side, we have nitrogen, and this side we have helium.

00:25 Initially, they're both at 100 kilopascals.

00:29 And initially, the volume of both are, i guess, you put a one here.

00:37 The volume of both of them are, is one cubic meter.

00:42 Um, let's see here.

00:45 It's not allowed, okay.

00:48 So this is, again, this moves, but doesn't allow any leakage of gas.

00:52 So you have everything over here stays over here and everything over here stays over here.

00:55 Now we can get the ideal gas constants for the, um, um, um, nitrogen and for the helium, the heat capacities for each of the gases.

01:07 And we're also going to wind up using, the universal ideal gas constant.

01:14 Now, we can figure out the mass of the nitrogen, because we know the pressure, the volume, and the temperature, and the hydro gas constant for the nitrogen in the first state.

01:23 So that winds up being 4 .772 kilograms.

01:27 We can also figure out the mass of the helium over here.

01:31 And that turns out to be 0 .8079 kilograms.

01:37 So a lot less helium over here, although we have the same volume.

01:41 Again, helium is a kind of a very low density.

01:50 So then we can say, okay, now if we take a control volume of the whole tank here, we know there's no work going in, it's rigid, there's no heat transfer coming in or out.

02:03 So we know the change in internal energy has to be zero.

02:06 And that change in internal energy is the change in internal energy in the nitrogen side, plus the change in internal energy in the helium side.

02:20 Now, we know then that when this whole thing comes into equilibrium, the temperatures on both sides are the same.

02:29 So tb2 and t -a -2 are the same, and i'll call that t -2.

02:32 And this gives us an equation for t -2 then.

02:36 And solving, we get that that is 57 .2 degrees celsius.

02:41 I don't know if i actually mentioned that we had the, temperatures in these sides is 80 degrees celsius and the nitrogen, 25 in the helium.

02:51 So we're, you know, in somewhere pretty close to being in the middle.

02:55 So what is what is 105, 105 over 2 is, you know, 52 .5.

03:05 So a little bit over halfway here.

03:12 So it's kind of surprising with the heat capacities this way.

03:15 But the fact is, is that we have a lot less mass of helium.

03:20 Now, we want the entropy generated.

03:25 So the entropy generated is just the change in entropy in each of the gases.

03:31 All right.

03:33 And so to get that, we need to, we know, we have the total volume here, right? we don't know that after this whole process takes place, we don't know where this piston is, right? so we don't know the specifics of the volume of each side.

03:49 I think we could find that.

03:52 But the easiest way to do this is to just say, we know the total volume.

03:55 The total volume doesn't change.

03:57 And the total volume is two cubic meters.

04:02 And we also know that this whole thing is at the same temperature.

04:12 And so if we know the number of moles of gas in here, it doesn't matter what type of gas it is.

04:22 If we know the number of moles, then we can use the ideal gas law.

04:25 To find the pressure in here.

04:29 And that's because, again, we're assuming that this thing has equilibrated.

04:34 This thing is maintaining the constant pressure here.

04:37 There's no net pressure pushing this either way.

04:40 So the pressure on either side is the same afterwards.

04:45 Now, to get that pressure, we need to, we use the ideal gas law, but we need first the number of moles of gas you have in here, total.

04:57 So the number of moles of gas we have in there is the number of moles of helium or in the helium and the number of modes of nitrogen.

05:03 And again, we know the masses of each of these things and we know the molar mass of each of them.

05:10 So for nitrogen, it is 28, yeah, 28.

05:20 And for helium it is four...

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