In this problem we want to compress a real gas, a non-ideal gas, carbon dioxide that
is initially at 175 degrees and 20 bar and we want to compress it to 275 bar and we want
to do this isothermally so it 175 degrees.
We want to determine what is the final volume, what's the internal energy change, entropy
change, the work done to compress the gas, and the total heat transfer during compression.
And we are going to model CO2 by the Peng Robinson Equation of State (EOS) and we are
going to use a spreadsheet that is described in a separate screencast to determine the
values and then we are going to use the values in the calculations.
So I want to show you know these values that I have read off the spreadsheet.
So the spreadsheet used the Peng-Robinson EOS to get the volume, internal energy and
the entropy.
We are interested in the final volume where we start with one cubic meter, a very large
container, and so what we need to determine since these are values of volume per mole,
we need to determine the number of moles in our system.
Since number of moles is not going to change.
So that is going to be the total volume of our system, divided by volume per mole when
we started.
So I'm going to convert this from cubic meters to cubic centimeters to be consistent with
the value from the spreadsheet.
So the number of moles initially, and the number of moles of course finally, is 551.
So we need that value to calculate the final volume in our container.
So the final volume, we should have called this volume 1 the total initial volume.
The total final volume would be the number of moles times the number of moles times the
volume per mole.
So 551 moles and then we read off the table, that we have 112.6 cubic cm per mole.
We are going to multiply these two, we get our final volume.
We have the final volume and then I converted it back to cubic meters and just wrote down
two significant figures.
Next we want to calculate delta U. Delta U total is going to be the number of moles times
U2 minus U1 where U2 and U1 are the kJ per mole that we obtain from the spreadsheet.
So let me substitute the values in.
So I've substituted the values in, and the important thing to notice is, if this was
an ideal gas, and for an ideal gas, isothermally, delta U would be zero.
So notice we have internal energy change for the real gas, we would not have this for an
ideal gas.
Next thing we want to calculate is the entropy change.
So we are going to do the same type calculation.
We are going to calculate the total entropy change as the number of moles, S2 minus S1.
Value of state functions we just need initial and final conditions, independent of path
to do this.
So let me substitute the numbers in here.
Again I have reduced the numbers to significant figures, converted this from joules to kJ.
The entropy is decreasing, kept it isothermal but decreased the volume, increased the pressure
so that should result in the decrease of entropy.
Now the last two parts we are going to have to make assumptions.
That's what the heat and work part is.
So to do this calculation we are going to have to assume it was done reversibly, otherwise
we have no way of determining it.
So what we are calculating then is the work reversible and the heat transfer reversible.
The heat transfer reversible is what we can calculate first because that's going to be
the temperature, since it's isothermal, times the total entropy change.
So of course the temperature has to be in absolute temperature, Kelvin, and the delta
S total that we just calculated.
And so Q is negative.
This says, as we expect, we have to remove heat.
We are doing work on the system when we compress it but since we are keeping it isothermal
we have to remove heat in order to satisfy the first law.
And so delta U total would be Q total + W total.
And we know now, delta U and Q so W total is just delta U total minus Q total.
So let's substitute those two values.
And so now we have the work total and you might round this to 5100 kJ.
Notice this is positive because we did work on the system in compressing it.
We have now calculated the work and the heat for the reversible.
So it is important to emphasis we assumed this was reversible in order to calculate
the work and heat otherwise we need more information.
Another thing to point out is when we have a non-ideal gas, modeling it, in this case
with the Peng Robinson EOS, then delta U is not zero for an isothermal process.