1.2: Additional Measurable Quantities

Introduction

In addition to the measurable sample specific quantities C_v and C_P there are a few other useful sample specific thermodynamic quantities that can be measured, tabulated and fit to functional forms for calculating properties of real samples. These are discussed below. What your should notice is that just like the heat capacities these are intensive properties and defined in terms of a derivative (change).

Isothermal Compressibility (\(\kappa_T\))

A very important property of a substance is how compressible it is. Gases are very compressible, so when subjected to high pressures, their volumes decrease significantly. Solids and liquids however are not as compressible. However, they are not entirely incompressible. High pressure will lead to a decrease in volume, even if it is only slight. Compressibility varies significantly from among substances even in the liquid and solid phases.

To quantify just how compressible substances are, it is necessary to define the property. The isothermal compressibility is defined by the fractional differential change in volume due to a change in pressure.

\[ \kappa_T \equiv - \dfrac{1}{V} \left( \dfrac{\partial V}{\partial P} \right)_T \label{compress} \]

The negative sign is important in order to keep the value of \(\kappa_T\) positive, since an increase in pressure will lead to a decrease in volume. The \(1/V\) term is needed to make the property intensive so that it can be tabulated in a useful manner.

Gases at high enough temperatures and low enough pressures (near ambient conditions for the gases you are familiar with) behave close to ideally. We can calculate \(\kappa_T\) for and idea gas by directly taking the derivative:

\[V = \frac{nRT}{P}\nonumber\]

So remembering that for an ideal gas \(\frac{nRT}{PV} = 1\):

\[\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = -\frac{1}{V}\left(\frac{\partial}{\partial P}\right)_T\frac{nRT}{P} = \frac{nRT}{P^2V} = P^{-1}\nonumber\]

So near ambient conditions \(\kappa_T\) for gases is (10⁵ Pa)^-1 = 10^-5 Pa^-1 ≈ 1 atm^-1 ≈ 1 bar^-1. The isothermal compressibility for liquids and solids are much smaller being in the range of 10^-5 atm^-1 for liquids and 10^-7 atm^-1 for solids.

Thermal expansion coefficient (\(\alpha\))

Another very important property of a substance is how its volume will respond to changes in temperature. Again, gases respond profoundly to changes in temperature (think Charles’ Law!) whereas solids and liquid will have more modest (but not negligible) responses to changes in temperature. (For example, If mercury or alcohol didn’t expand with increasing temperature, we wouldn’t be able to use those substances in thermometers.)

The definition of the thermal expansion coefficient (more accurately the isobaric thermal expansion coefficient) is

\[ \alpha \equiv \dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_P \label{expand} \]

As was the case with the compressibility factor, the \(1/V\) term is needed to make the property intensive, and thus able to be tabulated in a useful fashion. In the case of expansion, volume tends to increase with increasing temperature, so the partial derivative is positive.

Gases at high enough temperatures and low enough pressures (near ambient conditions for the gases you are familiar with) behave close to ideally. We can calculate \(\alpha\) for and idea gas by directly taking the derivative:

\[V = \frac{nRT}{P}\nonumber\]

So:

\[\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P = \frac{1}{V}\left(\frac{\partial}{\partial T}\right)_P\frac{nRT}{P} = \frac{nR}{PV} = T^{-1}\nonumber\]

Near room temperature this yields a value of (298 K)^-1 = 0.00336 K^-1. This apparently small number is actually large. For comparison with some solids and liquids see table \(\PageIndex{1}\).

Table \(\PageIndex{1}\): A sampling of thermal expansion coefficients at 20 ˚C (from Wikipedia.org, https://en.wikipedia.org/wiki/Therma...ious_materials)

Solids (K^-1)		Liquids (K^-1)
Al	69 x 10^-6	ethanol	0.000750
C (diamond)	3 x 10^-6	water	0.000207
Si	9 x 10^-6

Example \(\PageIndex{1}\): Using the reciprocal and cyclic partial derivative relations to find useful relationships among thermodynamic quantities

Derive an expression for

\[\dfrac{\alpha}{\kappa_T}. \label{e1} \]

in terms of derivatives of thermodynamic functions using the definitions in Equations \ref{compress} and \ref{expand}.

Solution

Substituting Equations \ref{compress} and \ref{expand} into the Equation \ref{e1}

\[\dfrac{\alpha}{\kappa_T}= \dfrac{\dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_P}{- \dfrac{1}{V} \left( \dfrac{\partial V}{\partial P} \right)_T} \nonumber \]

Simplifying (canceling the \(1/V\) terms and using the recipricol relation to invert the partial derivative in the denominator) yields

\[\dfrac{\alpha}{\kappa_T} = - \left( \dfrac{\partial V}{\partial T} \right)_P \left( \dfrac{\partial P}{\partial V} \right)_T \nonumber \]

Applying the cyclic relation to the product on the right allows it to be expressed as a single (and also measurable derivative):

\[ \dfrac{\alpha}{\kappa_T} = \left( \dfrac{\partial P}{\partial T} \right)_V \nonumber \]​​

Adiabatic compressibility (\(\kappa_S\))

In his seminal work, Philosophiae Naturalis Principia Mathematica (Newton, 1723), Isaac Newton (1643 - 1727)¹ calculated the speed of sound through air, assuming that sound was carried by isothermal compression waves. His calculated value of 949 m/s was about 15% smaller than experimental determinations. He accounted for the difference by pointing to “non-ideal effects”. But it turns out that his error, albeit an understandable one (since sound waves do not appear to change bulk air temperatures) was that the compression waves are adiabatic, rather than isothermal. As such, there are small temperature oscillations that occur due to the adiabatic compression followed by expansion of the gas carrying the sound waves. The oversight was correct by Pierre-Simon Laplace (1749 – 1827).²

LaPlace modeled the compression waves using the adiabatic compressibility, \(\kappa_S\) defined by

\[ \kappa_S =- \dfrac{1}{V} \left(\dfrac{\partial V}{\partial P} \right)_S \nonumber \]

Since the entropy is defined by

\[ dS = \dfrac{dq_{rev}}{T} \nonumber \]

it follows that any adiabatic pathway (\(dq = 0\)) is also isentropic (\(dS = 0\)), or proceeds at constant entropy.

Adiabatic pathways are also isentropic.

Using what we know about thermodynamics and the behaviors of partial derivatives we can quantitatively describe why realizing the compression is adiabatic increases the value calculated for the speed of sound. We will start with the formula for the speed of sound in a gas (see for example https://en.wikipedia.org/wiki/Speed_...ound#Equations for explanation of form):

\[v_{sound} = \sqrt{\dfrac{1}{\rho \, \kappa}} \nonumber \]

where \(\kappa\) is the compressibility. As noted above using the adiabatic compressibility yields a higher speed of sound. Thus \(\kappa_S \lt \kappa_T\), which we can show. 

Example \(\PageIndex{2}\): Showing that \(\kappa_S \lt \kappa_T\)

Show that \(\kappa_S \lt \kappa_T\) using known thermodynamic properties and derivative relations.

Solution

We begin by expanding the description of \(\kappa_S\) by using the cyclic partial derivative relation. Applying this, the adiabatic compressibility can be expressed

\[ \kappa_S =\dfrac{1}{V} \left(\dfrac{\partial V}{\partial S} \right)_P \left(\dfrac{\partial S}{\partial P} \right)_V \nonumber \]

followed by using the reciprocal relation:

\[ \kappa_S =\dfrac{1}{V} \dfrac{ \left(\dfrac{\partial S}{\partial P }\right)_V}{ \left(\dfrac{\partial S}{\partial V} \right)_P} \nonumber \]

Using a simple chain rule, the partial derivatives can be expanded to get something a little easier to evaluate:

\[ \kappa_S =\dfrac{1}{V} \dfrac{ \left(\dfrac{\partial S}{\partial T }\right)_V \left(\dfrac{\partial T}{\partial P }\right)_V }{ \left(\dfrac{\partial S}{\partial T} \right)_P \left(\dfrac{\partial T}{\partial V }\right)_P} \label{eq10} \]

The utility here is that

\[\left(\dfrac{\partial S}{\partial T }\right)_V = \dfrac{C_V}{T} \label{Note1} \]

\[\left(\dfrac{\partial S}{\partial T }\right)_P = \dfrac{C_P}{T} \label{Note2} \]

This means that Equation \ref{eq10} simplifies to

\[ \kappa_S = \dfrac{C_V}{C_P} \left( \dfrac{1}{V} \dfrac{ \left(\dfrac{\partial T}{\partial P }\right)_V }{ \left(\dfrac{\partial T}{\partial V }\right)_P} \right) \nonumber \]

Simplifying what is in the parenthesis yields

\[ \kappa_S = \dfrac{C_V}{C_P} \left( \dfrac{1}{V} \left(\dfrac{\partial T}{\partial P }\right)_V \left(\dfrac{\partial V}{\partial T }\right)_P \right) \nonumber \]

\[ \kappa_S = \dfrac{C_V}{C_P} \left( - \dfrac{1}{V} \left(\dfrac{\partial V}{\partial P }\right)_T \right) \nonumber \]

\[ \kappa_S = \dfrac{C_V}{C_P} \kappa_T \nonumber \]

Because it includes PV work, \(C_P\) is always bigger than \(C_V\), so \( \frac{C_V}{C_P} \lt 1\) making \(\kappa_S\) smaller than \(\kappa_T\).

Note

The above derivation also allows calculation of \(\kappa_S\) from readily measured quantities.

      References       

1. Isaac Newton, Wikipedia.org, https://en.wikipedia.org/wiki/Isaac_Newton.

2. O'Connor and Robertson, "Pierre-Simon Laplace", https://mathshistory.st-andrews.ac.u...phies/Laplace/.

Internal pressure (\(\pi_T\))

Another useful derivative quantity is the internal pressure. Going back to the expression for changes in internal energy that stems from assuming that \(U\) is a function of \(V\) and \(T\) (or \(U(V, T)\) for short)

\[ dU = \left( \dfrac{\partial U}{\partial V} \right)_TdV+ \left( \dfrac{\partial U}{\partial T} \right)_V dT \nonumber \]

one quickly recognizes one of the terms as the constant volume heat capacity, \(C_V\). And so the expression can be re-written

\[ dU = \left( \dfrac{\partial U}{\partial V} \right)_T dV + C_V dT \nonumber \]

But what about the first term? The partial derivative is a coefficient called the “internal pressure”, and given the symbol \(\pi_T\).

\[ \pi_T = \left( \dfrac{\partial U}{\partial V} \right)_T \nonumber \]

James Prescott Joule (1818-1889) recognized that \(\pi_T\) should have units of pressure (Energy/volume = pressure) and designed an experiment to measure it.

He immersed two copper spheres, A and B, connected by a stopcock. Sphere A is filled with a sample of gas while sphere B was evacuated. The idea was that when the stopcock was opened, the gas in sphere A would expand (\(\Delta V > 0\)) against the vacuum in sphere B (doing no work since \(P_{ext} = 0\). The change in the internal energy could be expressed

\[ dU = \pi_T dV + C_V dT \nonumber \]

But also, from the first law of thermodynamics

\[ dU = dq + dw \nonumber \]

Equating the two

\[ \pi_T dV + C_V dT = dq + dw \nonumber \]

and since \(dw = 0\)

\[ \pi_T dV + C_V dT = dq \nonumber \]

Joule concluded that \(dq = 0\) (and \(dT = 0\) as well) since he did not observe a temperature change in the water bath which could only have been caused by the metal spheres either absorbing or emitting heat. And because \(dV > 0\) for the gas that underwent the expansion into an open space, \(\pi_T\) must also be zero! In truth, the gas did undergo a temperature change, but it was too small to be detected within his experimental precision. An additional problem with this experiment is that unless the second container is of infinite volume the pressure the gas is expanding against steadily increases as gas moves to container B.

\(\pi_T\) is generally not zero and can be calculated from easily measurable quantities. This is shown below making use of what we know about differentials and Maxwell's relations.

The fundamental differential for U is:

\[dU = TdS - PdV \label{dU_fund}\]

We are looking for 

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

Dividing equation \(\ref{dU_fund}\) by dV and holding T constant yields:

\[\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial S}{\partial V} \right)_T - P\left(\frac{\partial V}{\partial V} \right)_T \nonumber\]

The last derivative is 1, so this simplifies to:

\[\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial S}{\partial V} \right)_T - P \nonumber\]

The derivative \(\left(\frac{\partial S}{\partial V} \right)_T\) is not easy to measure, so we want to express it in terms of more easily measurable quantities. Looking at the Maxwell relations, you will find the one from the dA fundamental equation \( \left( \dfrac{\partial P}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T \) suits our purpose. Making the substitution yields.

\[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial P}{\partial T} \right)_V -P \label{eq3} \]

Using what we showed in example \(\PageIndex{1}\), \(\left( \frac{\partial P}{\partial T} \right)_V = \frac{\alpha}{\kappa_T} \), we can substitute into equation \(\ref{eq3}\) to get a general expression in terms of easily measured quantities for the internal pressure of any substance (solid, liquid or gas):

\[ \pi_T = \left( \frac{\partial U}{\partial V} \right)_T = T \frac{\alpha}{\kappa_T}-P \nonumber \]

Note that since \(\alpha\) and \(\kappa_T\) are defined to be positive this implies that \(\pi_T\) is positive for "low" (near ambient for atmospheric gases) and negative at "high" pressure.

Example \(\PageIndex{3}\): Showing that \(\pi_T = 0\) for an ideal gas

Show that for an ideal gas \(\pi_T = \left( \frac{\partial U}{\partial V} \right)_T\) = 0.

Solution

For an ideal gas \(P = nRT/V\), so it is easy to show that

\[\left( \dfrac{\partial P}{\partial T} \right)_V = \dfrac{nR}{V} \label{eq4} \]

so combining Equations \ref{eq3} and \ref{eq4} together to get

\[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{nRT}{V} - P \label{eq5} \]

And since also because \(P = nRT/V\), then Equation \ref{eq5} simplifies to

\[ \left( \dfrac{\partial U}{\partial V} \right)_T = P -P = 0 \nonumber \]

So while Joule’s observation was consistent with limiting ideal behavior, his result was really an artifact of his experimental uncertainty masking what actually happened.

Example \(\PageIndex{4}\): Deriving \(\pi_T\) for a van der Waals gas

What is the internal pressure of a van der Waals gas?

Solution

For a van der Waals gas,

\[ P = \dfrac{nRT}{V-nb} - \dfrac{an^2}{V^2} \label{eqV1} \]

so

\[\left( \dfrac{\partial P}{\partial T} \right)_V = \dfrac{nR}{V-nb} \label{eqV2} \]

and

\[ \left( \dfrac{\partial U}{\partial V} \right)_T = T\dfrac{nR}{V-nb} - P \label{eqV3} \]

Substitution of the expression for \(P\) (Equation \ref{eqV1}) into this Equation \ref{eqV3}

\[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{an^2}{V^2} \nonumber \]

Summary of useful measurable derivatives

Table \(\PageIndex{2}\): The useful measurable derivatives.

Definitions	Alternative Expressions
\[ \kappa_T = - \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T\nonumber\]
\[ \alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P \nonumber \]
\[ \kappa_S =- \frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_S \nonumber \]
\[ \pi_T = \left( \frac{\partial U}{\partial V} \right)_T \nonumber \]	\[ \pi_T = \left( \frac{\partial U}{\partial V} \right)_T = T \frac{\alpha}{\kappa_T}-P \nonumber \]
\[ C_P = \left(\frac{\partial H}{\partial T}\right)_P \nonumber \]	\[ C_P = T\left(\frac{\partial S}{\partial T}\right)_P \nonumber \]
\[ C_V = \left(\frac{\partial U}{\partial T}\right)_V \nonumber \]	\[ C_V = T\left(\frac{\partial S}{\partial T}\right)_V \nonumber \]

Exercises

1. Starting with the fundamental differential expression for dH divide through by dT and hold P constant to get an expression for \(\left(\frac{\partial H}{\partial T}\right)_P\). Use this to show that both expressions for C_p in table \(\PageIndex{2}\) are valid.
2.  

Contributors

• Jonathan Gutow, UW Oshkosh

• Patrick Fleming, Cal State Eastbay