How to integrate a total differential along different paths?

Question

Starting with the ideal gas equation:

$V = RT/P$, the exact differential is ($R$ is a constant):

$$dV = \frac{R}{P}dT + \frac{-RT}{P^2}dP$$

The task is:

Show that $dV$ is an exact differential by integrating from $(P_1, T_1)$ to $(P_2, T_2)$ by two different paths: (i) $(P_1, T_1) \rightarrow (P_1, T_2)\rightarrow (P_2, T_2)$ and (ii) $(P_1, T_1) \rightarrow (P_2, T_1)\rightarrow (P_2, T_2)$. The result for both paths should of course be $(V_2 - V_1)$.

I am having a difficulty doing this. Methods to solve this require parameterisation, doing the entire path at once (?), but apparently this problem can be solved without this (?).

My feeling is that I am probably confusing what is a constant and what is a function.

_{source of the question: Thermodynamics of Natural Systems by Greg Anderson, page 10, question A1.1 (b)}

Answer 1

For example, if you chose the paths shown on the next figure :

Blue path :

From A to B ($T=T_1$=constant). $\int_{P_1}^{P_2} \frac{-RT_1}{P^2}dP=\frac{RT_1}{P_2}-\frac{RT_1}{P_1}$

From B to D ($P=P_2$=constant). $\int_{T_1}^{T_2} \frac{R}{P_2}dT=\frac{RT_2}{P_2}-\frac{RT_1}{P_2}$

$$\text{From A to D :}\quad\frac{RT_1}{P_2}-\frac{RT_1}{P_1}+\frac{RT_2}{P_2}-\frac{RT_1}{P_2} \tag 1$$

Red path :

From A to C ($P=P_1$=constant). $\int_{T_1}^{T_2} \frac{R}{P_1}dT=\frac{RT_2}{P_1}-\frac{RT_1}{P_1} $

From C to D ($T=T_2$=constant). $\int_{P_1}^{P_2} \frac{-RT_2}{P^2}dP=\frac{RT_2}{P_2}-\frac{RT_2}{P_1}$

$$\text{From A to D :}\quad\frac{RT_2}{P_1}-\frac{RT_1}{P_1}+\frac{RT_2}{P_2}-\frac{RT_2}{P_1} \tag 2$$

We see that the results $(1)$ and $(2)$ are equal $= \frac{RT_2}{P_2}-\frac{RT_1}{P_1}=V_2-V_1$ .

This is true for any points A, B, C, D.

So the property is true for any path made of linear segments such as P=constant and T=constant. This can be extended for any curved paths but the rigorous proof is more complicated even it can be intuitively understood. Obviously this proof is more complicated than with the simple equality of partial derivatives.

Answer 2

This is not an answer to the question, but a comment too long to be edited in the comments section.

The condition for $\quad M(x,y)dx+N(x,y)dy\quad$ to be an exact differential is : $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.$$ http://mathworld.wolfram.com/ExactDifferential.html

In the case of $\begin{cases} M(T,P)=\frac{R}{P} \\ N(T,P)=\frac{-RT}{P^2} \end{cases}$ $$ \frac{\partial M}{\partial P}=-\frac{R}{P^2}$$ $$ \frac{\partial N}{\partial T}=-\frac{R}{P^2}$$ They are equal. Thus $\quad\frac{R}{P}dT+\frac{-RT}{P^2}dP\quad$ is an exact differential.

Answer 3

The problem says "show that this is an exact differential by integrating along two different paths", not by looking at the mixed derivatives. The peculiar thing is that showing that you get the same value integrating along two different paths doesn't show that the differential is exact. You would have to show that you get the same result integrating along all possible paths!

Answer 4

You have $$dV = \frac{R}{P}dT - RT \frac{dP}{P^2}.$$

Along the first part of the first path, $P$ should be kept constant, so $dP=0$: $$ \int_{(P_1,T_1)}^{(P_1,T_2)} dV = \int_{T_1}^{T_2} \frac{R}{P_1} dT = \frac{R}{P_1} (T_2-T_1) $$

Along the second part of the first path, $T$ should be kept constant, so $dT=0$: $$ \int_{(P_1,T_2)}^{(P_2,T_2)} dV = \int_{P_1}^{P_2} (-RT_2) \frac{dP}{P^2} = (-RT_2) \left( \frac{-1}{P_2} - \frac{-1}{P_1} \right) $$

The total integral then becomes $$ \int_{\text{first path}} dV = \frac{R}{P_1} (T_2-T_1) + (RT_2) \left( \frac{1}{P_2} - \frac{1}{P_1} \right) \\ = \frac{RT_2}{P_1} - \frac{RT_1}{P_1} + \frac{RT_2}{P_2} - \frac{RT_2}{P_1} \\ = \frac{RT_2}{P_2} - \frac{RT_1}{P_1} \\ = V_2 - V_1. $$

Now do similar for the second path.