Chem 351 Help — Using Measurable Properties

Measurable Thermodynamic Properties

In this chapter, we have introduced six important thermodynamic derivatives:

\[ \boxed{C_p = \left( \frac{\partial H}{\partial T} \right)_p } \]

\[ \boxed{C_V = \left( \frac{\partial U}{\partial T} \right)_V } \]

\[ \boxed{\pi_T = \left( \frac{\partial U}{\partial V} \right)_T } \]

\[ \boxed{\mu_{JT} = \left( \frac{\partial T}{\partial p} \right)_H } \]

\[ \boxed{\alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_p } \]

\[ \boxed{\kappa_T = -\frac{1}{V} \left( \frac{\partial V}{\partial p} \right)_T } \]

Of these, the most useful experimentally are the heat capacities, the isobaric thermal expansivity, and the isothermal compressibility:

Worked examples

Worked example: Calculating an enthalpy change from a heat capacity

Calculate the change in enthalpy for a \(60.0\ \mathrm{g}\) piece of copper that is heated from \(19.2^\circ\mathrm{C}\) to \(87.4^\circ\mathrm{C}\).

From the Total Differential to \(\Delta H=nC_p\Delta T\)

For a constant-pressure change, the enthalpy depends on temperature through the constant-pressure heat capacity:

\[ dH = n \left( \frac{\partial H}{\partial T} \right)_p dT \]

Since the molar constant-pressure heat capacity is defined as

\[ C_p = \left( \frac{\partial H}{\partial T} \right)_p \]

we can write

\[ dH = nC_p\,dT \]

To calculate a finite enthalpy change, integrate from the initial temperature to the final temperature:

\[ \Delta H = \int_{T_1}^{T_2} nC_p\,dT \]

If \(C_p\) is independent of temperature over this range, it can be taken outside the integral:

\[ \Delta H = nC_p \int_{T_1}^{T_2} dT \]

\[ \boxed{ \Delta H = nC_p\Delta T } \]

The molar heat capacity of copper is

\[ C_p = 24.44\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

At constant pressure,

\[ \Delta H = nC_p\Delta T \]

First calculate the number of moles of copper:

\[ n = \frac{60.0\ \mathrm{g}} {63.546\ \mathrm{g\,mol^{-1}}} \]

\[ n = 0.944\ \mathrm{mol} \]

Next calculate the temperature change:

\[ \Delta T = 87.4^\circ\mathrm{C} - 19.2^\circ\mathrm{C} \]

\[ \Delta T = 68.2\ \mathrm{K} \]

Substituting into the heat-capacity expression:

\[ \Delta H = (0.944\ \mathrm{mol}) \left( 24.44\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) (68.2\ \mathrm{K}) \]

\[ \Delta H = 1.57\times10^3\ \mathrm{J} \]

\[ \boxed{ \Delta H = +1.57\ \mathrm{kJ} } \]

Physical interpretation: Because the copper is heated, energy flows into the sample and its enthalpy increases. The positive sign indicates that the process is endothermic.

Worked example: Calculating a volume change from thermal expansivity

Calculate the change in volume for the same \(60.0\ \mathrm{g}\) sample of copper when it is heated from \(19.2^\circ\mathrm{C}\) to \(87.4^\circ\mathrm{C}\).

For copper, use

\[ \alpha = 5.1\times10^{-5}\ \mathrm{K^{-1}} \]

and

\[ \rho = 8.96\ \mathrm{g\,cm^{-3}} \]

From the Definition of \(\alpha\) to \(\Delta V=\alpha V\Delta T\)

The isobaric thermal expansivity is defined as

\[ \alpha = \frac{1}{V} \left( \frac{\partial V} {\partial T} \right)_p \]

Rearranging gives

\[ dV = \alpha V\,dT \]

To calculate a finite volume change, integrate from the initial temperature to the final temperature:

\[ \Delta V = \int_{T_1}^{T_2} \alpha V\,dT \]

If \(\alpha\) is approximately independent of temperature and the volume change is small enough that \(V\) can be treated as approximately constant, then

\[ \Delta V = \alpha V \int_{T_1}^{T_2} dT \]

\[ \boxed{\Delta V = \alpha V\Delta T} \]

First calculate the initial volume of the copper sample from its mass and density:

\[ V = \frac{m}{\rho} = \frac{60.0\ \mathrm{g}} {8.96\ \mathrm{g\,cm^{-3}}} \]

\[ V = 6.70\ \mathrm{cm^3} \]

The temperature change is the same as in the previous example:

\[ \Delta T = 87.4^\circ\mathrm{C} - 19.2^\circ\mathrm{C} \]

\[ \Delta T = 68.2\ \mathrm{K} \]

Now calculate the volume change:

\[ \Delta V = \alpha V\Delta T \]

\[ \Delta V = (5.1\times10^{-5}\ \mathrm{K^{-1}}) (6.70\ \mathrm{cm^3}) (68.2\ \mathrm{K}) \]

\[ \Delta V = 2.3\times10^{-2}\ \mathrm{cm^3} \]

\[ \boxed{ \Delta V = +0.023\ \mathrm{cm^3} } \]

Physical interpretation: The volume change is positive because copper expands when heated. However, the change is very small compared with the original volume, illustrating that solids have much smaller thermal expansivities than gases.

Practice

Question 1
Which expression should be used to calculate the enthalpy change resulting from a temperature change at constant pressure?
\( \Delta H= \int_{T_1}^{T_2} \left( \frac{\partial H}{\partial T} \right)_p dT \)	\( \Delta H= \int_{p_1}^{p_2} \left( \frac{\partial H}{\partial p} \right)_T dp \)	\( \Delta H= \int_{V_1}^{V_2} \left( \frac{\partial H}{\partial V} \right)_T dV \)	\( \Delta H= \int_{T_1}^{T_2} \left( \frac{\partial T}{\partial H} \right)_p dT \)

Question 2
Which expression should be used to calculate the volume change resulting from a temperature change at constant pressure?
\( \Delta V= \int_{T_1}^{T_2} \left( \frac{\partial V}{\partial T} \right)_p dT \)	\( \Delta V= \int_{p_1}^{p_2} \left( \frac{\partial V}{\partial p} \right)_T dp \)	\( \Delta V= \int_{T_1}^{T_2} \left( \frac{\partial T}{\partial V} \right)_p dT \)	\( \Delta V= \int_{V_1}^{V_2} \left( \frac{\partial V}{\partial T} \right)_p dV \)

Question 3
Which expression should be used to calculate the volume change resulting from a pressure change at constant temperature?
\( \Delta V= \int_{p_1}^{p_2} \left( \frac{\partial V}{\partial p} \right)_T dp \)	\( \Delta V= \int_{T_1}^{T_2} \left( \frac{\partial V}{\partial T} \right)_p dT \)	\( \Delta V= \int_{p_1}^{p_2} \left( \frac{\partial p}{\partial V} \right)_T dp \)	\( \Delta V= \int_{V_1}^{V_2} \left( \frac{\partial V}{\partial p} \right)_T dV \)

Question 4
Suppose a thermodynamic property \(X\) depends on \(Y\) and \(Z\). Which expression correctly gives the change in \(X\) caused by changing \(Y\) while holding \(Z\) constant?
\( \Delta X= \int_{Y_1}^{Y_2} \left( \frac{\partial X}{\partial Y} \right)_Z dY \)	\( \Delta X= \int_{Y_1}^{Y_2} \left( \frac{\partial Y}{\partial X} \right)_Z dY \)	\( \Delta X= \int_{X_1}^{X_2} \left( \frac{\partial X}{\partial Y} \right)_Z dX \)	\( \Delta X= \int_{Z_1}^{Z_2} \left( \frac{\partial X}{\partial Z} \right)_Y dZ \)

Key points (one glance)

Thermodynamics makes extensive use of experimentally measurable properties to describe how state functions change.
Six important thermodynamic properties introduced in this chapter are \[ C_V,\qquad C_P,\qquad \pi_T,\qquad \mu_{JT},\qquad \alpha,\qquad \kappa_T \]
Of these, the most commonly measured and tabulated are \[ C_V,\qquad C_P,\qquad \alpha,\qquad \kappa_T \]
The constant-volume heat capacity is defined by \[ C_V = \left( \frac{\partial U} {\partial T} \right)_V \]
The constant-pressure heat capacity is defined by \[ C_P = \left( \frac{\partial H} {\partial T} \right)_p \]
The isobaric thermal expansivity is defined by \[ \alpha = \frac{1}{V} \left( \frac{\partial V} {\partial T} \right)_p \]
The isothermal compressibility is defined by \[ \kappa_T = -\frac{1}{V} \left( \frac{\partial V} {\partial p} \right)_T \]
If a thermodynamic property \(X\) depends on a variable \(Y\), then a finite change in \(X\) can often be calculated by integrating the corresponding partial derivative: \[ \Delta X = \int_{Y_1}^{Y_2} \left( \frac{\partial X} {\partial Y} \right)_Z dY \]
When the relevant coefficient is approximately constant over the range of interest, the integral often simplifies considerably.
For example, \[ \Delta H = \int_{T_1}^{T_2} nC_P\,dT \] becomes \[ \Delta H = nC_P\Delta T \] if \(C_P\) is independent of temperature.
Likewise, \[ \Delta V = \int_{T_1}^{T_2} \alpha V\,dT \] becomes \[ \Delta V = \alpha V\Delta T \] if \(\alpha\) is approximately constant and the volume change is small.
The power of thermodynamics lies in expressing difficult-to-measure derivatives in terms of measurable properties such as heat capacities, thermal expansivities, and compressibilities.

Big picture: Measurable properties such as \(C_P\), \(C_V\), \(\alpha\), and \(\kappa_T\) allow thermodynamic state functions to be connected directly to experiment. By integrating appropriate partial derivatives, changes in enthalpy, volume, internal energy, and other thermodynamic quantities can be calculated from experimentally accessible data.