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Dear Reader,

As my friend and fellow teacher is away in Germany I had to teach thermodynamics to the first years. I have to admit that I do not love all aspects of classical thermodynamics which is a bit of a misnomer it should be thermostatics as it deals mainly with static states.

One of the things I do not love about thermodynamics is that classical thermodynamics tells us nothing about how fast or the mechanism by which we go from A to B. All it tells us is the difference between A and B. The more modern thermodynamics which is based on the molecular motion can tell us more but I think that I will leave that some either some other day or some other person to write about.

As an incurable organic / inorganic chemist I love mechanism, mechanism explains to us how things happen. From that we can know what can and what can not happen.

One of the questions which I had to teach was the idea that we take an ideal gas and consider the heat capacity both under constant volume and constant pressure conditions.

For a monoatomic gas such as helium the heat capacities are

Cv = R 3/2 (constant volume)

Cp = R 5/2 (constant pressure)

If we were to assume that we have 2 grams of helium in a box which has a fixed volume of 1 litre then the box will have a pressure. The temperature of our box is 273 K. The formula mass (FW) of helium is 4 g mol-1

We can calculate this with PV = nRT

If we assume that R is equal to 8.3 J K-1 mol-1 then we can calculate P

n = M / FW = 0.5 moles

P = nRT / V = 0.5 mol x 8.3 J K-1 mol-1 x 273 K / 1 x 10-3 m3 = 1132950 Pa

OK so we have a start pressure of 1132950 Pa, or for those of us who do not like so many points before the decimal point we have about 1133 kPa or circa 1.133 MPa. This is about 10 bar.

Now if we heat the box up to 100 oC (373 K) then the energy of the helium will increase.

As Cv = ΔU / ΔT

We can rearrange our equation to give us

Cv ΔT = ΔU = 8.3 x 3/2 x 100 K = 1245 J mol-1

As we have half a mole then we have to put in 622.5 J of energy if the volume is going to stay the same. The pressure will increase we can predict it with the following equation.

P2 / P1 = T2 / T1

My calculations give us a value of 1547950 Pa for the final value.

Now lets do it again with constant pressure

As Cp = ΔH / ΔT = 5/2 R

If the pressure stays the same then the amount of energy required will be higher at 1037.5 J.

We might for a moment ask why is the energy higher for the fixed pressure case than it is for the fixed volume.

When the gas is heated up at a fixed pressure then it has to expand to continue to satisfy the equation PV = nRT

We can calculate the volume increase

V2/V1 = T2/T1

So the box has to increase by 366.3 ml to a volume of 1.3663 litres or 1.3663 x 10-3 m3

Now E = P ΔV

For fun lets do the dimensional analysis

Joules = N m-2 x m3 = N m

Now E = P ΔV = 1132950 Pa x 0.3663 x 10-3 m3 = 415 J

Now 622.5 J + 415 J = 1037.5 J

I am hoping that you can see that the difference between the two energies which need to be added to the gas to increase the temperature. I hope to add some diagrams shortly.


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