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The Isothermal Bulk Modulus: Instructor's Guide

Main Ideas

  • Partial Derivatives
  • Total Differentials
  • Chain Rule Diagrams
  • Thermodynamic Variables
  • Sackur-Tetrode Entropy Equation

Students' Task

Estimated Time: 50 minutes (includes 20 min. wrap-up)

  1. Given the expressions found in the Isothermal Bulk Modulus handout, students in small groups are asked to find the equations for two partials provided.
  2. Students will use chain rule diagrams to keep from getting lost in the complex expressions and will use the cyclic chain rule to simplify solving the second partial derivative.

Prerequisite Knowledge

  • Familiarity with total differentials
  • Familiarity with the cyclic chain rule
  • Familiarity with partial derivatives and their interpretations
  • Familiarity with chain rule diagrams is useful


Activity: Introduction

This activity works well as a follower of the Calculating a Total Differential activity. Many of the concepts of taking a total differential and using a chain rule diagram to keep track of differentials will apply to this activity as well. Students are placed into small groups and asked to find expressions for the differentials $$-V \left(\frac{\partial p}{\partial V}\right)_{T}$$ and $$-V \left(\frac{\partial p}{\partial V}\right)_S .$$

Activity: Student Conversations

We use this activity early in the course to give students practice working with partial derivatives, and to demonstrate mathematically that the “thing held constant” really does make a difference. Finding the isothermal bulk modulus is pretty easy. Finding the adiabatic bulk modulus requires figuring out how to hold entropy fixed. There are two distinct approaches that one can use for this problem. One is to apply rules for partial derivatives and changes of variables, which is the traditional approach. The other (which I lean towards) is to use total differentials alone (plus algebra) to solve this problem.

The total differential approach requires that one look for an equation that looks like: \[ dp = A dV + B dS \] where $A$ and $B$ are quantities to solve for. Once we have an equation that looks like this, we can immediately identify \[ A = \left(\frac{\partial p}{\partial V}\right)_S \] and we are done. We find an equation in this form by constructing total differentials of the equations provided and algebraically eliminating the differentials we don't want in the final equation. The only requirement is that we never divide by an infinitesimal quantity.

Activity: Wrap-up

  • For this problem, the use of chain rule diagrams will help to not get lost in the math. Although plugging and chugging with the three expressions given can, in principle, lead to the correct answers, using chain rule diagrams is less time consuming and more graceful.
  • An important obstacle to overcome in this activity is when working with partial derivatives, we can only assume that the variable notated as being held constant is the sole constant term; any other variables can not just be considered constant when taking the derivative of an expression. Using the cyclic chain rule allows the bypassing of challenging derivatives with the substitution of two, more reasonable partials. However, a note of warning should be given to students about overusing the cyclic chain rule, particularly that for most problems using the cyclic chain rule more than once will lead to dead-ends and trivial statements.


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