Thermodynamics, manipulating partial derivatives

[MexChemE]

Hello PF! It's been a while since I last posted here. I have come across a problem in my textbook, which asks me to find expressions for V as a function of T and P, starting from the coefficients of thermal expansion and compressibility.
$$\alpha = \frac{1}{V} \left(\frac{\partial V}{\partial T} \right)_P$$
$$\beta = -\frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_T$$
I've already solved the problem, I separated the differentials and integrated, then cleared for V. Now, here's where I have trouble. This is how I manipulated the differentials:
$$\alpha \ dT= \frac{1}{V} \ dV$$
$$\beta \ dP= -\frac{1}{V} \ dV$$
Both my professor, and Castellan's PChem text say this is correct, however, neither my professor nor the book explain why the separation of a partial derivative works in this case. Math professors have always said we can't "break" a partial derivative the same way we do for a regular derivative. If anyone could offer some insight about this case, it would be very helpful. Thanks!

 

[sweet springs]

$$\alpha dT=\frac{dV}{V}$$ under the condition p=const or dp=0
$$\beta dP=\frac{-dV}{V}$$ under the condition T=const or dT=0
You need these conditions.

 

[Chestermiller]

If V is a function V(P,T) of P and T, then
$$dV=\left(\frac{\partial V}{\partial P}\right)_TdP+\left(\frac{\partial V}{\partial T}\right)_PdT$$

Chet

 

[MexChemE]

Thank you both! I understand now, I thought they were just magically turning the partial derivative into a regular one. I wasn't aware of how both coefficients relate to the differential of V. We haven't covered partial derivatives or differentials of multivariable functions in Calc III, my professor is going a bit slowly. But we have started to use them in Thermodynamics I, so everything I know about the subject right now is from what I have read on my own.

Oh, and I'm sorry you had to move my thread, I thought it could be posted in the physics section since I wasn't asking for help on solving the problem itself, just a doubt with the math involved, but I'll be more careful next time.

 

[paranormal]

Hello there! It's great to see you actively solving problems and seeking clarification on concepts. Let me provide some insight on why the separation of partial derivatives works in this case.

Firstly, it's important to understand that partial derivatives represent the rate of change of a function with respect to one variable, while holding all other variables constant. In other words, they tell us how a function changes when one variable changes, while keeping the others fixed.

In thermodynamics, we often deal with functions that depend on multiple variables, such as temperature (T), pressure (P), and volume (V). In your problem, you are asked to find an expression for volume (V) as a function of temperature (T) and pressure (P). This means that the volume (V) is dependent on both temperature (T) and pressure (P), and we need to take into account how it changes when either of these variables changes.

Now, let's look at the expressions for thermal expansion coefficient (\alpha) and compressibility (\beta) that you provided. These are defined as the ratio of the change in volume to the original volume, when temperature (T) or pressure (P) changes, respectively. In other words, they are partial derivatives of volume (V) with respect to temperature (T) and pressure (P). Therefore, we can write:

\alpha = \frac{1}{V} \left(\frac{\partial V}{\partial T} \right)_P
\beta = \frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_T

Now, let's look at the manipulation of differentials that you did. You multiplied both sides of the equations by the differentials (dT and dP) and rearranged them to get:

\alpha \ dT= \frac{1}{V} \ dV
\beta \ dP= \frac{1}{V} \ dV

This is a valid manipulation, as we can treat the differentials (dT and dP) as independent variables, while keeping the partial derivatives as constants. This is because the differentials represent small changes in temperature and pressure, which can be treated as independent variables in this case. Therefore, we can separate the partial derivative from the differential and treat it like a regular derivative.

I hope this explanation helps you understand why the separation of partial derivatives works in this case. Keep up the good work and don