4.1

So in this worked example, we’re going to look at a concept that is being marketed by MDI in France and was demonstrated in Peugeot 208 at the 2014 Paris Auto Show. Now we’re asked to calculate the available energy, or the changing Gibbs energy, in a 100 litre tank of compressed air at 248 bars, if it’s released isothermally through an air motor in a car running at a constant speed along a level road. So as in all the other examples that we’ve looked at, I’m going to start by drawing a picture and summarising the information on the picture. So we have some compressed air. And we’ve got a volume of 100 litres of it at a pressure of 248 bars.

71.2

And I’m going to call this state 1. And this is when it’s inside the tank. So here’s my tank. I’m going to let it escape. And so now it will be expanded air. And its new pressure, P2, will be at one bar atmospheric pressure. And it’s got an unknown new volume. And this is going to be state 2. And I’m going to use the air as my system. So I’m going to draw my system boundary around the tank but also include the expanded air.

120.4

So that’s my system boundary. I’m going to apply the first law of thermodynamics between state 1 and state 2 and then also use the second law of thermodynamics to solve this question. So applying the first law– so apply the first law of thermodynamics.

146.9

A standard expression for this is Q2 minus Q1. They’re the heat transfer terms. Plus the work terms, W2 minus W1, plus the terms due to energy carried by any mass flow across the boundary, so it’s Emass1 minus Emass2.

172.4

And all of these are going to be equal to the change in the internal energy of the system plus the change in the kinetic energy of the system plus the change in the potential energy of the system. So not all of these terms are going to have values in this particular problem. We’re told that the car is moving at constant speed along a level road. So that means we can set this change to 0. So this is 0 due to a level road. It’s moving at constant speed. So this term here will go to 0. So this is 0 due to constant speed.

216.8

And we’re also told this expansion is isothermal, so that means there’s no change in temperature. And so the internal energy must also be– or the change in internal energy must also be 0. So this is 0 due to isothermal process.

239.3

And then no air is lost here. There’s no transfer of mass across my system boundary. And so that means that this term here must also go to 0. So this is 0 due to no mass transfer.

261

And so that just leaves us with these terms here, the heat transfer and the work terms. And so we can write this down as Q2 minus Q1 is equal to W1 minus W2. Or we could express that as the change in the heat transfer is equal to the change in the work.

290

And for an ideal gas, we can express the work done in terms of the pressure times the volume. So we can say that for an ideal gas, PV equals mRT. So it’s pressure times volume times the mass times the gas constant times the temperature in degrees Kelvin. And so we can extend this expression here and say dQ is equal to dW is equal to pdV. And then that will give mRT upon V times when we substitute the ideal gas equation. And now let’s apply the second law. So we’ll hold on to that for a moment and say, apply second law for an ideal reversible system.

366.1

And the idea reversal bit allows me to express it with an equal sign here and say that the change in entropy is equal to the heat transfer change divided by the temperature at which it occurs. And so now we can take this little bit here, where mRT and dV upon V is equal to dQ. And so we can substitute that in here. And we get mR over V times dV. I’m going to call that equation 1. We’re going to hang on to that. And I’d like to clear the board and continue the second part of the problem. So now let’s move on to the second part of this example, which is to calculate the change in Gibbs energy.

411.9

And so first of all, we can write that by definition, Gibbs energy– I’m going to use a simple G for Gibbs energy– is equal to enthalpy minus the temperature times the entropy. And we can expand this enthalpy term as U, the internal energy, plus pressure times volume, and then minus temperature times entropy. But we want the change in Gibbs energy. And so change in Gibbs energy– and we can express that as dG, change in Gibbs energy, is going to be equal to dU plus d times change in pressure times volume, minus the change in temperature times entropy. And this one here is going to be 0, as it was in the first law, because it’s an isothermal process.

492.8

And this one also will go to 0, and again, actually due to it being an isothermal process. So due to isothermal process– now this one’s perhaps a little less obvious. We can substitute here and say this is dPV will be equal to– using the ideal gas equation, this will be equal to dmRT, and all of these are constants. And so if you look at the change in them, it’s going to be equal to 0. So let’s maybe just annotate that all constants. And so that means that we’re just left with these terms here. And so we can say, and dG is equal to minus TdS.

549.2

And then we can use this expression up here, because we can say that ds is equal to mR over Vdv. And so substituting (1), we’re going to end up with dG is equal to minus mRT upon V times dv. And so this is the small change in Gibbs energy. We’re interested in the change over this whole process from 1 through to 2, and so we need to integrate over that process. So now integrating over process of air escaping– and so we’re going to end up with the integral between 2 and 1 of dG. And again, the integral between 1 and 2 of this thing here. Simpl take the minus sign outside, mRT upon V times dv.

625.4

And that implies that this will simply become G2 minus G1. And the mRT is a constant, so we don’t need to do anything to that. And then the integral of 1 upon V is ln, natural logarithm, of V. And so if we substitute between the limits, it will be V2 upon V1. And we don’t know what the V1 is, but we can go back and use the ideal gas equation, because this is air. And so we can substitute this back again and get minus– oh, I forgot my minus here as well– minus P1V1 ln P2 upon P1, remembering that temperature is constant. So we don’t need to worry about that when we substitute our ideal gas equation.

687.1

So we know all these values. And so we can substitute in appropriate values. So P1 is 248 bars. So that’s 10 to the 5 if we do it in Pascals, multiplied by the volume here, which is 100 litres. So we express that in cubic metres. That’s 0.1. And all that is going to be times ln he ratio of the pressures between these , which is going to be 248 upon 1. And if you get your calculator out and do that calculation, it comes out at 13.67 times 10 to the 6 joules. And so that’s the amount of energy available, the chang in Gibbs energy, when we allow this compressed air to escape from the tank.

743.8

And this is the most optimistic estimate, because we used the equality version of the second law. So we’re pushing the second law right to the absolute limit. This value here is about 1/3 of the energy available from a litre of diesel. So actually in terms of propulsion of cars, it’s not a huge amount of energy. But MDI claimed 90% efficiency for their air motors, which is a lot better than most internal combustion engines. And on this basis, they claim a range of around 220 kilometres on a tank of compressed air. So that makes it a viable vehicle.