1
Microscopic Origin of Thermodynamics

Problem 1.1 Using the phase rule, justify whether the following thermodynamic systems are adequately defined:

1. a pure liquid of known mass at given temperature and density;
2. a pure liquid of known mass at given density and pressure;
3. a two‐phase system at given temperature, pressure, and the mole fractions of all chemical species in one phase;
4. an ionic liquid containing two types of cations and one type of anions at given temperature, pressure, and the ionic composition.

Solution

The phase rule indicates that, for a bulk system with chemical species and phases without chemical reaction, intensive variables are needed to define a thermodynamic state. In addition, one extensive variable is required to define the system size.

1. Yes, here and , thus . Temperature and density are independent intensive variables, mass is an extensive quantity.
2. The specification could be problematic because pressure and density are conjugated variables. For example, liquid water exhibits a maximum density at 1 atm and 4 °C, implying that there could be two temperatures for one density at the same pressure.
3. No, the system is over specified. then . But temperature, pressure, and the compositions of one phase constitute independent intensive variables.
4. Here (two ion pairs), and . The intensive variables are well defined but the system size is unknown.

1. Problem 1.2 Based on the fundamental thermodynamic relations, show

   Solution

   First, write the constant‐pressure heat capacity in terms of entropy

   Taking a partial derivative with respect to at constant and leads to

   From the Maxwell relation

   we then have

2. Problem 1.3 Acoustic velocity, or the speed of sound, in a fluid can be derived from the first law of thermodynamics for steady‐state flow processes

   in combination with the mass‐balance equation

   Here, is the specific enthalpy of the fluid, is the mass density, and is the velocity of the elastic wave, i.e., the sound speed.

   1. Derive the following expression from the 1st law and mass balance
   2. Show that the partial derivative at constant entropy can be expressed in terms of measurable quantities where is the ratio of the constant‐pressure and constant‐volume heat capacities of the fluid, is the isothermal compressibility
   3. Predict the speed of sound in the atmosphere at 25 C. Assuming atmosphere is an ideal gas with molar heat capacity  J/(mol K) and molecular weight  g/mol.

   Solution

   1. Combing the 1st law with the fundamental equation leads to, at constant entropy, where . The expression for the sound speed is obtained by dividing the above equation by that from the mass balance, ,
   2. First, we rewrite the partial derivative at constant entropy in terms the molar volume V: To make a connection with measurable quantities, we use the following identities: and Combining these equations leads to where we have used the following equations Therefore, the partial derivative at constant entropy becomes and where is the ratio of the constant‐pressure and constant‐volume heat capacities of the fluid, and is the isothermal compressibility
   3. For an ideal gas, the compressibility is and where is the gas constant. At 1 atm and 25 C, the mass density of atmosphere is where we have used the average molecular weight of air  g/mol. Substituting along with the molar heat capacity  J/(mol K) into the equation for the sound speed yields  m/s.
3. Problem 1.4 Consider a one‐dimensional model for a thermodynamic system containing identical particles. Assume that the internal energy is completely determined by temperature , system total length , and particle number . Show that the fundamental equation of thermodynamics can be written as

   where temperature, line tension (i.e., negative of the one‐dimensional pressure), and chemical pressure are defined as, respectively,

   What are the analogs of the Maxwell relations for this one‐dimensional system?

   Solution

   We may derive the fundamental equation of thermodynamics from , which is a linear homogeneous function

   In analogy with the conventional fundamental equation for three‐dimensional systems, we can derive the Maxwell relations from the second‐order differentiations of the internal energy:

4. Problem 1.5 Two‐dimensional models are often used to describe the thermodynamics of molecular adsorption from a gas or a liquid solution on planar surfaces. For a one‐component two‐dimensional system, the fundamental equation is given by

   where stands for surface pressure. Derive the Gibbs adsorption isotherm:

   Can you extend the Gibbs adsorption isotherm to multicomponent systems? What is the surface pressure if there is no adsorption?

   Solution

   For this system, the integrated form of the fundamental equation is

   In comparison with its differential form, we can derive the Gibbs–Duhem equation for the two‐dimensional system

   At constant temperature, , thus

   For multicomponent systems, the Gibbs–Duhem equation becomes

   In this case, the Gibbs adsorption isotherm becomes

   If there is no adsorption, and the surface pressure vanishes, i.e., .

5. Problem 1.6 Consider gas adsorption on a planar surface at temperature . Assume that the gas phase is ideal and that the adsorption can be described by Henry's law

   where is the number of gas molecules at the surface, is the surface area, and is Henry's constant. Show that the surface pressure is given by

   where is Boltzmann's constant.

   Solution

   For an ideal gas in the bulk, the chemical potential depends on pressure

   where is the chemical potential of the ideal gas at system temperature and a reference pressure  atm, and take the same units.

   The combination of the Gibbs isotherm with Henry's law predicts

   With the boundary condition at , integrating the above equation leads to

   Using Henry's law again, we obtain an analog of the ideal‐gas law for two dimensional systems,

6. Problem 1.7 Consider molecular adsorption from a liquid solution on a planar surface of area and temperature . Assume that the liquid phase is an ideal solution and that the adsorption of the solute molecules can be described by Henry's law

   where is the moles of solute molecules at the surface, is Henry's constant, and stands for the molar concentration of the solute molecules in the bulk solution. Show that the surface pressure is given by

   where is the gas constant.

   Solution

   The chemical potential of solute molecules in an ideal solution can be written as

   where is the chemical potential of the solute at a reference state of system temperature and  mol/L. With Henry's law , the Gibbs isotherm predicts

   With the boundary condition at , integrating the above equation leads to

   Using Henry's law again, we obtain a two‐dimensional analog of the ideal‐gas law for the surface pressure

7. Problem 1.8 Consider ideal‐gas adsorption on a planar surface such that gas molecules at the surface follows a two‐dimensional equation of state

   where is the surface area per molecule, and is the surface area occupied by each molecule. Show that the adsorption isotherm follows the Volmer isotherm

   where stands for the surface coverage, and is Henry's constant defined by as . How would you extend the Volmer equation to adsorption on a planar surface from an ideal solution?

   Solution

   For an ideal gas at constant temperature , the chemical potential varies with the bulk pressure

   The two‐dimensional equation of state predicts that the surface pressure varies with the surface area per molecule

   Substituting these two equations into the Gibbs isotherm (Problem 1.5) yields

   Integrating the above equation leads to

   where is an integration constant. Replacing with gives

   (A)

   The integration constant can be fixed with the boundary condition as ,

   Substituting into Eq. (A) leads to the Volmer equation

   For adsorption from an ideal solution, we replace with the molar concentration of the solute

   which leads to

   where as .

8. Problem 1.9 Consider ideal‐gas adsorption on a planar surface such that gas molecules at the surface follows...