• Blog Stats

    • 71,430 hits
  • Archives

  • Enter your email address to subscribe to this blog and receive notifications of new posts by email.

    Join 146 other followers

  • Copyright notice

    This blog entry and all other text on this blog is copyrighted, you are free to read it, discuss it with friends, co-workers and anyone else who will pay attention.

    If you want to cite this blog article or quote from it in a not for profit website or blog then please feel free to do so as long as you provide a link back to this blog article.

    If as a school teacher or university teacher you wish to use content from my blog for the education of students then you may do so as long as the teaching materials produced from my blogged writings are not distributed for profit to others. Also at University level I ask that you provide a link to my blog to the students.

    If you want to quote from this blog in an academic paper published in an academic journal then please contact me before you submit your paper to enable us to discuss the matter.

    If you wish to reuse my text in a way where you will be making a profit (however small) please contact me before you do so, and we can discuss the licensing of the content.

    If you want to contact me then please do so by e-mailing me at Chalmers University of Technology, I am quite easy to find there as I am the only person with the surname “foreman” working at Chalmers. An alternative method of contacting me is to leave a comment on a blog article. If you do not know which one to comment on then just pick one at random, please include your email in the comment so I can contact you.

Thermodynamics

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.

Advertisements

Go on, Have your say !

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: