Problems involving exact differentials : Exercises

Introduction

A differential,

$\textrm{d}f = C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}y$ , is exact if it has the following property:

$\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x}$ Furhtermore, if the differential is exact then we can write

$\textrm{d}f = \left( \frac{\partial f}{\partial x}\right)_y\textrm{d}x + \left( \frac{\partial f}{\partial y}\right)_x\textrm{d}y$ . The exercises that follow involve learning how to test differentials for exactness and how to solve partial differential equations involving exact differentials. The exercises below should be completed as follows:

If you click on the first box below you will find a complete worked solution to the question it asks.
The second box in the first section contains a set of comprehension questions that should be answered by looking through the worked solution that was given to the first problem.
The boxes in the second question contain problems for you to work on. Only the final solution is given when you click on the question to reveal an answer.
The two questions in the final section involve applying what you have learnt about exact differentials to thermodynamics.

Example problems

Click on the problems to reveal the solution

Problem 1

Test the differential $\textrm{d}u = (y-x^2) \textrm{d}x + (x+y^2) \textrm{d}y$ for exactness and find the function $u(x,y)$ .

An exact differential,

$\textrm{d}f=C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}x$ , has the following important properties:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ Furthermore,

$\textrm{d}f = \left(\frac{\partial f}{\partial x}\right)_y \textrm{d}x + \left(\frac{\partial f}{\partial y}\right)_x \textrm{d}y$ for an exact differential. Two final useful results can be obtained by comparing coefficients of

$\textrm{d}x$ and

$\textrm{d}y$ in

$\textrm{d}f=C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}y$ and

$\textrm{d}f = \left(\frac{\partial f}{\partial x}\right)_y \textrm{d}x + \left(\frac{\partial f}{\partial y}\right)_x \textrm{d}y$ . These two results being of course:

$\left( \frac{\partial f}{\partial x} \right)_y = C_1(x,y) \qquad \textrm{and} \qquad \left( \frac{\partial f}{\partial y} \right)_x = C_2(x,y)$ We can then insert these results into

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$ to obtain:

$\left(\frac{\partial C_1}{\partial y}\right)_x = \left(\frac{\partial C_2}{\partial x}\right)_y$ In the example given in the question

$C_1 = (y-x^2) \qquad \rightarrow \qquad \left(\frac{\partial C_1}{\partial y}\right)_x = 1$

$C_2 = (x+y^2) \qquad \rightarrow \qquad \left(\frac{\partial C_2}{\partial x}\right)_y = 1$

$\textrm{d}u$ is thefore an exact differential because

$\left(\frac{\partial C_1}{\partial y}\right)_x = \left(\frac{\partial C_2}{\partial x}\right)_y$ . We can find

$u$ using indefinite integrals. We know that:

$\left( \frac{\partial u}{\partial x} \right)_y = y - x^2 \qquad \qquad \left( \frac{\partial u}{\partial y} \right)_x = x + y^2$ We can thus integrate along the

$x$ axis (i.e. keeping

$y$ fixed) and get the following.

$\int (y - x^2)_y \textrm{d}x = yx - \frac{1}{3}x^3 + k(y)$ where

$k(y)$ is some unknown function of

$y$ . Similarly:

$\int (x + y^2)_x \textrm{d}y = yx + \frac{1}{3}y^3 + c(x)$ For the above two functions to be consistent the functions

$c(x)$ and

$k(y)$ must be equal to:

$c(x) = -\frac{1}{3}x^3 \qquad k(y) = \frac{1}{3}y^3$ Hence:

$u(x,y) = \frac{1}{3} y^3 +yx - \frac{1}{3}x^3 + C$

Use the information in the worked example above to answer the following questions:

For the exact differential $\textrm{d}f = C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}y$ how are the functions $C_1(x,y)$ and $C_2(x,y)$ related to the partial derivatives of the function $f$ ?

Explain why $\left(\frac{\partial C_1}{\partial y}\right)_x = \left(\frac{\partial C_2}{\partial x}\right)_y$ must hold for the differential $\textrm{d}f = C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}y$ to be exact.

Explain what we mean when we state that we integrate along the $x$ axis and why the additive constant that we obtain when we do so depends on $y$ .

What results from the earlier parts of the proof must the solution, $u(x,y)=\frac{1}{3} y^3 +yx - \frac{1}{3}x^3 + C$ , be consistent with.

For an exact differential, $\textrm{d}f = C_1(x,y)\textrm{d}x + C_2(x,y)\textrm{d}y$ , $C_1(x,y)$ is the partial derivative of $f$ with respect to $x$ and $C_2$ is the partial derivative of $f$ with respect to $y$ . In other words, $\left( \frac{\partial f}{\partial x} \right)_y = C_1(x,y)$ and $\left( \frac{\partial f}{\partial y} \right)_x = C_2(x,y)$
For an exact differential $\left(\frac{\partial C_1}{\partial y}\right)_x = \left(\frac{\partial C_2}{\partial x}\right)_y$ because, as discussed in the previous part, $C_1(x,y)$ and $C_2(x,y)$ must be equal to the first partial derivatives of $f(x,y)$ with respect to $x$ and $y$ respectively. In other words, because $C_1(x,y) = \left( \frac{\partial f}{\partial x} \right)_y$ $\left(\frac{\partial C_1}{\partial y}\right)_x = \left(\frac{\partial^2 f}{\partial y \partial x}\right)$ . In addition, $C_2(x,y) = \left( \frac{\partial f}{\partial y} \right)_x$ so $\left(\frac{\partial C_2}{\partial x}\right)_x = \left(\frac{\partial^2 f}{\partial x \partial f}\right)$ . These two second-derivative, cross terms must be equal as $\textrm{d}f$ is an exact differential.
By integrating along $x$ we mean that we integrate along a path that is parallel to the $x$ axis. Now obviously there is are an infinite number of paths that run paralle to the $x$ axis - these paths differ in the value that $y$ takes. The integral under each of these different paths will be different and as such the constant term depends on $y$ .
$u(x,y)=\frac{1}{3} y^3 +yx - \frac{1}{3}x^3 + C$ must be consistent with the integral obtained by integrating along a 1D path that runs parallel to the $x$ axis, $\int (y - x^2)_y \textrm{d}x = yx - \frac{1}{3}x^3 + k(y)$ and the result obtained by integrating along a 1D path that runs paralle to the $y$ axis $\int (x + y^2)_x \textrm{d}y = yx + \frac{1}{3}y^3 + c(x)$ . To arrive at the final result we solve for the two functions $k(y)$ and $c(x)$ by setting these two integrals equal.

Problem 2

Test if the differential $\textrm{d}u = 2xy \textrm{d}x + (x^2 + 3y^2) \textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = x^2 y + y^3 + C$

Test if the differential $\textrm{d}u = 2xy \textrm{d}x + x^2 \textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = x^2 y + C$

Test if the differential $\textrm{d}u = (3x + 3) \textrm{d}x + (2y - 2)\textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = \frac{3}{2} x^2 + 3x + y^2 - 2y + C$

Test if the differential $\textrm{d}u = (2xy - 9x^2)\textrm{d}x + (2y + x^2 + 1) \textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = x^2y -3x^3 + y^2 + y + C$

Test if the differential $\textrm{d}u = (2xy^2 + 4)\textrm{d}x - 2(3 - x^2y)\textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = x^2y^2 + 4x - 6y + C$

Test if the differential $\textrm{d}u = \left(\frac{2xy}{x^2 + 1} - 2x\right)\textrm{d}x - \left[ 2 - \ln(x^2 + 1) \right]\textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = y\ln(x^2 + 1) - 2y - x^2 + C$

Test if the differential $\textrm{d}u = (3y^3 e^{3xy} - 1 )\textrm{d}x + ( 2ye^{3xy} + 3xy^2 e^{3xy})\textrm{d}y$ is exact. If it is find the function $u(x,y)$ .

$u = y^2 e^{3xy} - x + C$

Problem 3

Consider the equation of state of an ideal gas $PV=Nk_BT$ . Clearly $V$ can be expressed as a function of $P$ and $T$ , $V(P,T)$ . Show that $\textrm{d}V$ is an exact differential.

Rearanging the equation of state for the ideal gas that we were given in the question we find that:

$V = \frac{N k_B T}{P}$ We can thus write the differential

$\textrm{d}V$ as:

$\textrm{d}V = \left( \frac{\partial V}{\partial P}\right)_T \textrm{d}P + \left(\frac{\partial V}{\partial T}\right)_P \textrm{d}T = -\frac{N k_B T}{P^2} \textrm{d}P + \frac{N k_B}{P} \textrm{d}T$ The coefficients of

$P$ and

$T$ in this expression are thus

$\begin{aligned} C_1 & = -\frac{N k_B T}{P^2} \qquad \rightarrow \qquad \left( \frac{\partial C_1}{\partial T}\right)_P = -\frac{N k_B}{P^2} \\ C_2 & = \frac{N k_B}{P} \qquad \rightarrow \qquad \left( \frac{\partial C_2}{\partial P}\right)_T = -\frac{N k_B}{P^2} \end{aligned}$ The above two partial derivatives are equal so

$\textrm{d}V$ is thus an exact differential. This agrees with our physical intuition. Volume is a state function so whenever the system is moved between two particular thermodynamic states the the total change in volume is the same. The path that is taken between these two states does not matter.

Consider the equation of state for the ideal gas once more and determine if $-P\textrm{d}V$ an exact differential?

Taking the differential we obtained for

$\textrm{d}V$ in the previous part and multiplying it by

$-P$ gives:

$-P\textrm{d}V = \frac{N k_B T}{P} \textrm{d}P - N k_B \textrm{d}T$ We thus have:

$\begin{aligned} C_1 & = \frac{N k_B T}{P} \qquad \rightarrow \qquad \left( \frac{\partial C_1}{\partial T}\right)_P = \frac{N k_B}{P} \\ C_2 & = N k_B \qquad \rightarrow \qquad \left( \frac{\partial C_2}{\partial P}\right)_T = 0 \end{aligned}$ These two derivatives are not equal so

$-P\textrm{d}V$ is not an exact differential. In actual fact the integral of

$-P\textrm{d}V$ is equal to the total amount of

$PV$ -work that is done during the course of the transition between our thermodynamic states. The work done during the transition does depend on the path we take between the two thermodynamic states. For instance the ammount of work done when the system is not allowed to exchange heat with its surroundings will be diffent to the ammount of work that will be done when the system is also allowed to exchange heat with the surroundings. This result is thus in accordance with physical intutition once more.

Problems involving exact differentials : Exercises

Introduction

Example problems

Problem 1

Problem 2

Problem 3

Contact Details