The Properties of Gases – Equation of State

The equation of state of a gas relates its macroscopic variables pressure (\(p\)), volume (\(V\)), and temperature (\(T\)). For an ideal gas, the equation of state is the Ideal Gas Law:

\[ pV = nRT \]

Where \(p\) is the pressure of the gas, \(V\) is the volume of the gas, \(T\) is the temperature of the gas, \(R\) is the universal gas constant, \(n\) is the number of moles of the gas.

 

Empirical Gas Laws

The ideal gas is based on several empirical gas laws:

Boyle’s Law: At constant temperature (\(T\)) (isothermal process) and number of moles (\(n\)), the pressure and volume are inversely proportional:

\[ pV = \text{constant} \]

Charles’s Law: At constant pressure (\(p\)) (isobaric process) and number of moles (\(n\)), the volume and temperature are directly proportional:

\[ V \propto T \implies  \frac{V}{T} = \frac{V'}{T'} \]

Gay-Lussac’s Law: At constant volume (isochoric process) and number of moles (\(n\)), the pressure and temperature are directly proportional

\[ p \propto T \implies \frac{p}{T} = \frac{p'}{T'}\]

Avogadro’s Principle: At constant temperature and pressure, the volume of a gas is proportional to the number of moles:

\[ V \propto n \]

 

Partial Derivatives and the Ideal Gas Law

In thermodynamics, partial derivatives are commonly used to express how a thermodynamic property changes with one variable while keeping other variables constant. In this case, we will apply partial derivatives to the ideal gas law.

The ideal gas law is given by:

\[ p V = nRT \]

We can rearrange this equation to solve for pressure:

\[ p = \frac{nRT}{V} \]

This allows us to express pressure as a function of volume, temperature, and the number of moles of gas:

\[ p(n, T, V) \]

Next, we will explore how pressure changes with temperature and volume, assuming other variables are held constant.

To calculate the partial derivative of pressure with respect to temperature while keeping the volume and number of moles constant, we take the following derivative:

\[ \left( \frac{\partial p}{\partial T} \right)_{V,n} \]

Substituting the ideal gas law into this expression, we have:

\[ \left( \frac{\partial}{\partial T} \left( \frac{nRT}{V} \right) \right)_{V,n} = \frac{nR}{V} \]

This tells us that the pressure increases linearly with temperature when the volume and number of moles are held constant. Specifically, this result represents the rate of change of pressure per unit increase in temperature.

Now, let's find the partial derivative of pressure with respect to volume while keeping the temperature and number of moles constant:

\[ \left( \frac{\partial p}{\partial V} \right)_{T,n} \]

Substituting the ideal gas law gives:

\[ \left( \frac{\partial}{\partial V} \left( \frac{nRT}{V} \right) \right)_{T,n} = - \frac{nRT}{V^2} \]

This result shows that pressure decreases with increasing volume. The negative sign indicates that this is an inverse relationship, meaning that as volume increases, pressure decreases when temperature and the number of moles is constant.

 

Thermal Expansion Coefficient and Isothermal Compressibility

Thermodynamic properties such as the thermal expansion coefficient and isothermal compressibility can also be expressed using partial derivatives.

Thermal Expansion Coefficient (\( \alpha \)) measures the fractional change in volume per unit change in temperature at constant pressure:

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

Isothermal Compressibility (\( \kappa_T \)) measures the fractional change in volume per unit change in pressure at constant temperature:

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

Both of these properties are important for describing how a substance responds to changes in external conditions, such as pressure and temperature.

 

Example: Applying the Ideal Gas Law to a Specific Scenario

Let's consider a sample of nitrogen gas (N2) at room temperature (300 K) and atmospheric pressure (1 atm). Using the ideal gas law:

\[ pV = nRT \]

Assume we have 1 mole of nitrogen gas. The gas constant \(R = 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\). Rearranging the ideal gas law for volume, we get:

\[ V = \frac{nRT}{p} = \frac{(1 \, \text{mol})(0.0821)(300 \, \text{K})}{1 \, \text{atm}} = 24.63 \, \text{L} \]

This tells us the volume occupied by 1 mole of nitrogen gas at room temperature and atmospheric pressure. We can now calculate the thermal expansion coefficient and isothermal compressibility for this gas under these conditions using the expressions derived above.

 

Real Gases – Deviations from Ideal Behavior

Real gases deviate from ideal behavior, especially at high pressures or low temperatures, due to:

  • Finite size of gas molecules i.e. excluded volume is not negligible.
  • Intermolecular forces - Attractive/Repulsive forces are not negligible.
  • High pressure: repulsion dominates due to short-range interaction.
  • Low temperature: attraction dominates due to long-range interaction.

Lennard-Jones Potential

The Lennard-Jones potential describes the interaction between two neutral molecules (or atoms) as a function of distance between them. It accounts for both attractive forces (like van der Waals attraction) and repulsive forces (Pauli repulsion) through the following formula:

\[ E_p(r) = 4 \varepsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^6 \right] \]

  • r - Distance between the particles
  • \(\sigma\) - Distance at which the potential \(E_p\) = 0 (characteristic distance of the potential)
  • \(\varepsilon\) - Depth of the potential well (determines the strength of the attraction)

The Lennard-Jones potential has two major parts:

  • Repulsive term \(\left( \frac{\sigma}{r} \right)^{12}\): Dominates at short distances due to Pauli exclusion principle, representing the steep repulsion.
  • Attractive term \(\left( \frac{\sigma}{r} \right)^6\): Dominates at longer distances due to van der Waals (London dispersion) forces.

The graph of the Lennard-Jones Potential shows the potential energy curve with the repulsive and attractive regions. The minimum of the curve represents the equilibrium distance where the attractive and repulsive forces balance each other out. At very short distances, the repulsion dominates due to overlapping electron clouds (Pauli exclusion), and at longer distances, the attractive van der Waals forces dominate.

 

Compression Factor Z

The compression factor ($Z$) is a measure of the deviation of real gas behavior from ideal gas behavior.

  • at lower pressures: the distance between particles is larger, so attractive forces dominate. Real gases are easier to compress.
  • at higher pressures: the distance between particles is shorter, so repulsive forces dominate. Real gases are harder to compress.

The compression factor is given by the ratio of the molar volume of a real gas (V_m,RG) to the molar volume of an ideal gas (V_m,IG):

\[ Z = \frac{V_{m,RG}}{V_{m,IG}} \]

Since for an ideal gas, we know:

\[ V_{m,IG} = \frac{RT}{P} \]

Thus, for an ideal gas, Z becomes:

\[ Z = \frac{P V_m}{RT} \]

For an ideal gas, Z = 1. For a real gas:

  • If Z > 1, repulsive forces are dominant, and the molar volume increases with increasing pressure.
  • If Z < 1, attractive forces dominate, and the molar volume decreases with increasing pressure.

The Virial Equation of State

One way to describe real gas behavior more accurately is to use the virial equation of state. It expands the compression factor as a power series of $\frac{1}{Vm}$:

\[ Z = \frac{PV}{nRT} = 1 + \frac{1}{V_m}B(T) + \left( \frac{1}{V_m} \right)^2 C(T) + \left( \frac{1}{V_m} \right)^3 D(T) + \cdots \]

Here, B(T), C(T), D(T), etc., are the virial coefficients, which depend on temperature and account for intermolecular forces. The higher-order terms become significant at higher densities.

The graphical representation shows that the compressibility factor Z changes depending on the gas and the conditions, as indicated by the curves for different gases (O2, CO2, H2) at 400 K.

 

van der Waals equation of state

The van der Waals equation corrects the Ideal Gas Law to account for intermolecular forces and the volume occupied by gas molecules. This equation corrects two major assumptions of the ideal gas law: that gas molecules do not interact with each other and that they do not occupy any volume.

Formulation:

The ideal gas law is expressed as:

\[ p V = n R T \]

In reality, real gases deviate from this behavior. The van der Waals equation modifies the volume and pressure terms as follows:

\[ (p + \frac{n^2 a}{V^2})(V - n b) = n R T \]
or
\[ \left( p + \frac{a}{V_m^2} \right)(V_m - b) = RT \]
or
\[ p = \frac{RT}{V_m - b} - \frac{a}{V_m^2} \]

Where:

  • \(a\) accounts for intermolecular attractive forces
  • \(b\) accounts for the finite volume occupied by the gas molecules (excluded volume)

Volume Correction - Excluded Volume

Real gas molecules are not point particles and occupy a finite volume. The correction for this volume is expressed as:

\[ V \rightarrow V - nb \]

The excluded volume, \( b \), is the volume inaccessible to other molecules due to the finite size of each molecule.

The volume of one molecule is:

\[ V_{molecule} = \frac{4}{3} \pi r^3 \]

and the excluded volume for two molecules is:

\[ V_{excluded} = \frac{4}{3} \pi (2r)^3 = 8V_{molecule} \]

Where \( r \) is the effective radius of the molecule.

Therefore, the total inaccessible volume is:

\[ b = 4 V_{molecule} N_A \]

The van der Waals equation provides a more accurate description of real gases compared to the ideal gas law, especially at high pressures and low temperatures. The modifications to the pressure and volume terms help account for molecular interactions and the finite size of gas molecules.

It is particularly useful in understanding the behavior of gases that are near the point of condensation, as it better predicts the critical temperature and pressure where gases liquefy.

 

Derivation of the Critical Constants

The critical constants (\(T_c\), \(p_c\), \(V_c\)) describe the conditions at the critical point, where the distinction between liquid and gas phases disappears. These constants are derived from the van der Waals equation.

At the critical point, the slope of the isotherm is zero, meaning:

  • First derivative of pressure with respect to volume:

\[ \left( \frac{\partial p}{\partial V_m} \right)_T = 0 \]

  • Second derivative of pressure with respect to volume:

\[ \left( \frac{\partial^2 p}{\partial V_m^2} \right)_T = 0 \]

Rearranging the van der Waals equation to express pressure as a function of molar volume and temperature and taking the first and second derivatives of \(p\) with respect to \(V_m\):

\[ p = \frac{RT}{V_m - b} - \frac{a}{V_m^2} \]

\[ \frac{\partial p}{\partial V_m} = - \frac{RT}{(V_m - b)^2} + \frac{2a}{V_m^3} = 0 \implies \frac{2a}{V_m^3} = \frac{RT}{(V_m-b)^2} \]

\[ \frac{\partial^2 p}{\partial V_m^2} = \frac{2RT}{(V_m - b)^3} - \frac{6a}{V_m^4} = 0 \implies \frac{6a}{V_m^4} = \frac{2RT}{(V_m - b)^3} \]

Solving these as a system of equation gives the critical volume $V_{m,c}$:

\[ \frac{V_m}{3} = \frac{V_m - b}{2} \implies V_{m,c} = 3b \]

Substituting \(V_{m,c} \) into the van der Waals equation allows us to find the critical pressure \(p_{m,c}\), while substituing \(V_{m,c} \) in the first derivative allows us to find the critical temperature $T_c$:

  1. Critical temperature:
    \[ T_c = \frac{8a}{27Rb} \]
  2. Critical pressure:
    \[ p_c = \frac{a}{27b^2} \]