# 4.6: Useful Definitions and Relationships


In this chapter (and in the previous chapter), several useful definitions have been stated.

Toolbox of useful Relationships

The following “measurable quantities” have been defined:

• Heat Capacities: $C_V \equiv \left( \dfrac{\partial U}{\partial T} \right)_V$ and$C_p \equiv \left( \dfrac{\partial H}{\partial T} \right)_p$
• Coefficient of Thermal Expansion: $\alpha \equiv \left( \dfrac{\partial V}{\partial T} \right)_p$ or $\left( \dfrac{\partial V}{\partial T} \right)_p = V \alpha$
• Isothermal Compressibility: $\kappa_T \equiv - \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_T$ or $\left( \dfrac{\partial V}{\partial p} \right)_T = -V \kappa _T$

The following relation has been derived:

$\dfrac{ \alpha}{\kappa_T} = \left( \dfrac{\partial p}{\partial T} \right)_V$

And the following relationships were given without proof (yet!):

$\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V - p$

and

$\left( \dfrac{\partial H}{\partial p} \right)_T = - T \left( \dfrac{\partial V}{\partial T} \right)_p - p$

Together, these relationships and definitions make a powerful set of tools that can be used to derive a number of very useful expressions.

Example $$\PageIndex{1}$$: Expanding Thermodynamic Function

Derive an expression for $$\left( \dfrac{\partial H}{\partial V} \right)_T$$ in terms of measurable quantities.

Solution 1:

Begin by using the total differential of $$H(p, T)$$:

$dH = \left( \dfrac{\partial H}{\partial p} \right)_T dp + \left( \dfrac{\partial H}{\partial T} \right)_p dT$

Divide by $$dV$$ and constrain to constant $$T$$ (to generate the partial of interest on the left):

$\left.\dfrac{dH}{dV} \right\rvert_{T}= \left( \dfrac{\partial H}{\partial p} \right)_T \left.\dfrac{dp}{dV} \right\rvert_{T} + \cancelto{0}{\left( \dfrac{\partial H}{\partial T} \right)_p \left.\dfrac{dT}{dV} \right\rvert_{T}}$

The last term on the right will vanish (since $$dT = 0$$ for constant $$T$$). After converting to partial derivatives

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = \left( \dfrac{\partial H}{\partial p} \right)_T \left(\dfrac{\partial p}{\partial V} \right)_{T} \label{eq5}$

This result is simply a demonstration of the “chain rule” on partial derivatives! But now we are getting somewhere. We can now substitute for $$\left(\dfrac{\partial H}{\partial V} \right)_{T}$$ using our “toolbox of useful relationships”:

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = \left[ -T \left(\dfrac{\partial V}{\partial T} \right)_{p} +V \right] \left(\dfrac{\partial p}{\partial V} \right)_{T}$

Using the distributive property of multiplication, this expression becomes

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = -T \left(\dfrac{\partial V}{\partial T} \right)_{p}\left(\dfrac{\partial p}{\partial V} \right)_{T} + V \left(\dfrac{\partial p}{\partial V} \right)_{T} \label{eq7}$

Using the cyclic permutation rule (Transformation Type II), the middle term of Equation \ref{eq7} can be simplified

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = T \left(\dfrac{\partial p}{\partial T} \right)_{V} + V \left(\dfrac{\partial p}{\partial V} \right)_{T}$

And now all of the partial derivatives on the right can be expressed in terms of $$\alpha$$ and $$\kappa_T$$ (along with $$T$$ and $$V$$, which are also “measurable properties”.

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = T \dfrac{\alpha}{\kappa_T} + V \dfrac{1}{-V \kappa_T}$

or

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = \dfrac{1}{\kappa_T} ( T \alpha -1)$

Example $$\PageIndex{2}$$: Isothermal Compression

Calculate $$\Delta H$$ for the isothermal compression of ethanol which will decrease the molar volume by $$0.010\, L/mol$$ at 300 K. (For ethanol, $$\alpha = 1.1 \times 10^{-3 }K^{-1}$$ and $$\kappa_T = 7.9 \times 10^{-5} atm^{-1}$$).

Solution

Integrating the total differential of $$H$$ at constant temperature results in

$\Delta H = \left(\dfrac{\partial H}{\partial V} \right)_{T} \Delta V$

From Example $$\PageIndex{1}$$, we know that

$\left(\dfrac{\partial H}{\partial V} \right)_{T} = \dfrac{1}{\kappa_T} ( T \alpha -1)$

so

$\Delta H = \left [ \dfrac{1}{ 7.9 \times 10^{-5} atm^{-1}} \left( (300 \,K) (1.1 \times 10^{-3 }K^{-1}) -1 \right) \right] ( - 0.010\, L/mol )$

$\Delta H = \left( 84.81 \, \dfrac{\cancel{atm\,L}}{mol}\right) \underbrace{\left(\dfrac{8.314\,J}{0.8206\, \cancel{atm\,L}}\right)}_{\text{conversion factor}} = 9590 \, J/mol$

Contributors

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

4.6: Useful Definitions and Relationships is shared under a not declared license and was authored, remixed, and/or curated by Patrick Fleming.