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} m^{3} = 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 ^{o}C (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.

P_{2} / P_{1} = T_{2} / T_{1}

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

V_{2}/V_{1} = T_{2}/T_{1}

So the box has to increase by 366.3 ml to a volume of 1.3663 litres or 1.3663 x 10^{-3} m^{3}

Now E = P ΔV

For fun lets do the dimensional analysis

Joules = N m^{-2} x m^{3} = N m

Now E = P ΔV = 1132950 Pa x 0.3663 x 10^{-3} m^{3} = 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.

Filed under: Chemistry, thermodynamics | Tagged: classical thermodynamics |

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