7.1 Thermodynamic Properties, Ideal Gas, and State Equations

Thermodynamic State and Fundamental Properties

Thermodynamics in chemical engineering deals with the relationship between heat, work, and the properties of matter. For the NCEES FE Chemical exam, a solid grasp of thermodynamic definitions and mathematical relationships is critical. Properties are categorized as either intensive (independent of the amount of substance, such as temperature $T$, pressure $P$, and specific volume $v$) or extensive (dependent on system mass, such as total volume $V$, internal energy $U$, and entropy $S$). State functions (such as $U$, $H$, $S$, $A$, and $G$) depend only on the current state of the system, not on the path taken to reach it. In contrast, heat ($Q$) and work ($W$) are path functions; their values depend on the specific thermodynamic path of the process.

The fundamental thermodynamic relations combine the First and Second Laws. For a closed system of constant composition undergoing a reversible process, the differential change in internal energy is:

$$dU = TdS - PdV$$

From this definition, other key thermodynamic properties are defined mathematically in the NCEES FE Reference Handbook:

• Enthalpy ($H$): $H = U + PV$. Differentiating yields the fundamental equation: $$dH = TdS + VdP$$
• Helmholtz Free Energy ($A$): $A = U - TS$. Differentiating yields: $$dA = -SdT - PdV$$
• Gibbs Free Energy ($G$): $G = H - TS$. Differentiating yields: $$dG = -SdT + VdP$$

These differential relations lead directly to the Maxwell relations via the mathematical properties of exact differentials. For example, since $dG$ is an exact differential, the partial derivatives of its coefficients must be equal:

$$\left(\frac{\partial V}{\partial T}\right)_P = -\left(\frac{\partial S}{\partial P}\right)_T$$

These relations allow engineers to express unmeasurable properties like entropy changes in terms of measurable properties like temperature, pressure, and volume.

The Ideal Gas Law and Compressibility Factor

At sufficiently low pressures and high temperatures, all gases behave ideally. The Ideal Gas Law is written as:

$$PV = nRT \quad \text{or} \quad Pv = RT$$

where $P$ is absolute pressure, $V$ is total volume, $n$ is the number of moles, $v$ is molar volume ($V/n$), $T$ is absolute temperature, and $R$ is the universal gas constant. Candidates must be highly proficient in unit conversions using $R$. In SI units, $R = 8.314 \text{ J/(mol}\cdot\text{K)} = 8.314 \text{ kPa}\cdot\text{m}^3\text{/(kmol}\cdot\text{K)}$. In USCS units, $R = 10.73 \text{ psia}\cdot\text{ft}^3\text{/(lbmol}\cdot\text{R)} = 1545 \text{ ft}\cdot\text{lbf/(lbmol}\cdot\text{R)}$. Always ensure pressure and temperature are in absolute units (e.g., Kelvin or Rankine, not Celsius or Fahrenheit).

For real gases, intermolecular forces and finite molecular volume cause deviations from ideal behavior. The degree of deviation is quantified by the compressibility factor ($Z$):

$$Z = \frac{Pv}{RT}$$

For an ideal gas, $Z = 1$. When $Z < 1$, attractive forces dominate, making the gas easier to compress than an ideal gas. When $Z > 1$, repulsive forces dominate, making the gas harder to compress.

According to the three-parameter Theorem of Corresponding States, all gases exhibit the same deviation from ideal behavior when compared at the same reduced temperature ($T_r$) and reduced pressure ($P_r$):

$$T_r = \frac{T}{T_c}, \quad P_r = \frac{P}{P_c}$$

where $T_c$ and $P_c$ are the critical temperature and pressure of the gas. The compressibility factor $Z$ can be read from generalized compressibility charts in the FE Reference Handbook as a function of $T_r$ and $P_r$. If molar volume is sought, the pseudo-reduced specific volume ($v'_r$) is used:

$$v'_r = \frac{v}{R T_c / P_c}$$

Real Gas Equations of State

When high precision is required, semi-empirical cubic equations of state (EOS) are utilized. The two most common cubic equations on the FE Chemical exam are the van der Waals and Redlich-Kwong equations of state.

The van der Waals (vdW) equation accounts for intermolecular attraction (via parameter $a$) and molecular volume (via parameter $b$):

$$\left(P + \frac{a}{v^2}\right)(v - b) = RT$$

where $v$ is the molar volume. The parameters $a$ and $b$ are determined from the critical properties by forcing the first and second derivatives of pressure with respect to volume to zero at the critical point:

$$a = \frac{27 R^2 T_c^2}{64 P_c}, \quad b = \frac{R T_c}{8 P_c}$$

The Redlich-Kwong (RK) equation improves upon the vdW equation by introducing a temperature dependence in the attractive term:

$$P = \frac{RT}{v - b} - \frac{a}{T^{0.5} v (v + b)}$$

The parameters $a$ and $b$ for the RK equation are defined in the FE Reference Handbook as:

$$a = 0.42748 \frac{R^2 T_c^{2.5}}{P_c}, \quad b = 0.08664 \frac{R T_c}{P_c}$$

Worked Example: Real Gas Pressure Calculation

Determine the pressure exerted by $1.0 \text{ kmol}$ of methane ($CH_4$) confined to a volume of $0.15 \text{ m}^3$ at $200 \text{ K}$ using the van der Waals equation of state. Methane critical properties are $T_c = 190.6 \text{ K}$ and $P_c = 46.0 \text{ bar} = 4600 \text{ kPa}$. The universal gas constant is $R = 8.314 \text{ kPa}\cdot\text{m}^3\text{/(kmol}\cdot\text{K)}$.

Step 1: Calculate parameters $a$ and $b$.

$$a = \frac{27 (8.314)^2 (190.6)^2}{64 (4600)} = \frac{27 (69.12) (36328.36)}{294400} = \frac{67784347}{294400} \approx 230.2 \text{ kPa}\cdot(\text{m}^3/\text{kmol})^2$$ $$b = \frac{(8.314) (190.6)}{8 (4600)} = \frac{1584.6}{36800} \approx 0.0431 \text{ m}^3/\text{kmol}$$

Step 2: Calculate the molar volume $v$.

$$v = \frac{V}{n} = \frac{0.15 \text{ m}^3}{1.0 \text{ kmol}} = 0.15 \text{ m}^3/\text{kmol}$$

Step 3: Solve for pressure $P$.

$$P = \frac{RT}{v - b} - \frac{a}{v^2}$$ $$P = \frac{(8.314) (200)}{0.15 - 0.0431} - \frac{230.2}{(0.15)^2}$$ $$P = \frac{1662.8}{0.1069} - \frac{230.2}{0.0225}$$ $$P = 15555 \text{ kPa} - 10231 \text{ kPa} = 5324 \text{ kPa} = 53.2 \text{ bar}$$

Comparing this to the ideal gas law ($P = RT/v = (8.314 \times 200)/0.15 = 11085 \text{ kPa}$), the real gas pressure is significantly lower due to the strong attractive forces represented by parameter $a$, which is expected since $T = 200 \text{ K}$ is very close to the critical temperature ($190.6 \text{ K}$).