I have a question which is somewhat physics-y but it's still a maths question at heart, I hope someone here can help.

Basically, here is the problem, I get to this intermediate stage in a thermo proof:

$$ dH = T \bigg( \frac{\partial S}{\partial T} \bigg)_P dT $$

The next stage just skips to $$ \bigg( \frac{\partial H}{\partial T} \bigg)_P = T \bigg( \frac{\partial S}{\partial T} \bigg) _P $$

Did they just divide by dT? I thought that, that was not something that is even possible (in terms of proper maths). I've tried applying the product rule but then I end up in the precarious position of trying to to figure out what the derivative of dT is wrt to T, can someone please guide me through how you get from the first equation to the second in formal math terms please!

Thank you.

  • $\begingroup$ You should be careful about your notation. The notation for derivatives in thermodynamics is slightly different than in mathematics in general. $\endgroup$ – Michael McGovern Apr 2 '17 at 20:40
  • $\begingroup$ @MichaelMcGovern Which part in particular, is it the brackets? Should they be replaced with a modulus type notation? $\endgroup$ – MathsIsHard Apr 2 '17 at 20:41
  • $\begingroup$ No. The brackets are ok and are the most common use in physics. The brackets are just a very carefull notation so that you can see what is considered constant in the partial differential $\endgroup$ – Rafael Wagner Apr 2 '17 at 20:42
  • $\begingroup$ The biggest problem is from total derivative to an partial differential equation. $\endgroup$ – Rafael Wagner Apr 2 '17 at 20:43
  • $\begingroup$ @RafaelWagner Ah, that may be because the dH came from dH = TdS + vdP. At constant P, this reduced to dH = TdS, a substitution was then made for TdS, so H is a multivariable function. $\endgroup$ – MathsIsHard Apr 2 '17 at 20:46

Let $H = H(T,p)$ just for an example. Then we say that the total differential of H is given by

$$\mathrm dH = \frac{\partial H}{\partial T}\mathrm dT + \frac{\partial H}{\partial p}\mathrm dp $$

Now let $p$ be constant then you don't have a variation in pressure and it becomes

$$\mathrm dH = \frac{\partial H}{\partial T}\mathrm dT$$

To be precise in order to others can understand what is constant physicians wright

$$\mathrm dH = \left(\frac{\partial H}{\partial T}\right)_{p}\mathrm dT\tag{1}$$

Your first equation says that

$$\mathrm dH = T \left( \frac{\partial S}{\partial T} \right)_P \mathrm dT\tag{2}$$

See $(1)$ and $(2)$ together then you'll have the equality you want. An important question is: Why the $H$ have a total differential well defined in the first place? This is a really important physical and mathematical question.

  • $\begingroup$ Thank you! As for your final question...I am not sure. $\endgroup$ – MathsIsHard Apr 2 '17 at 20:55
  • $\begingroup$ Is it because Enthalpy can be completely defined with two variables, because the two other variables are functions of all else? $\endgroup$ – MathsIsHard Apr 2 '17 at 20:55
  • $\begingroup$ @MathsIsHard What he mean is the first law of thermodynamic $\endgroup$ – quallenjäger Apr 2 '17 at 21:10

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