Ideal and Real Gas Equations of State

Real gas equation of state

An ideal gas obeys Pv = RT, but a real gas deviates from this law, most strongly near saturation and at high pressure. The deviation is measured by the compressibility factor Z = \frac{Pv}{RT}, which equals 1 for an ideal gas. For most real gases at moderate conditions Z is below 1 because attractive forces dominate, and at high pressure Z can exceed 1 because the finite molecular volume dominates. Once Z is known, the actual specific volume is v = Z\,\frac{RT}{P}, so the ideal-gas estimate is corrected by multiplying it by Z.

The van der Waals equation is the standard two-constant model: \left(P + \frac{a}{v^2}\right)(v - b) = RT. The constant a accounts for intermolecular attraction, which raises the effective pressure, and b accounts for the finite molecular volume, which reduces the free volume. When v is per mole, a is expressed in units such as kPa·(m³/kmol)² and b in m³/kmol, and R must be in matching units such as kJ/(kmol·K). Temperature is then found by direct substitution, T = \frac{(P + a/v^2)(v-b)}{R}, and it is compared with the ideal-gas value T = \frac{Pv}{R}. A short check: if a is set to zero and b is also zero, the van der Waals result reduces to the ideal-gas result.

Thermodynamic property relations of a real gas follow from its equation of state. For enthalpy, dh = c_P\,dT + \left[v - T\left(\frac{\partial v}{\partial T}\right)_P\right]dP, so \left(\frac{\partial h}{\partial P}\right)_T = v - T\left(\frac{\partial v}{\partial T}\right)_P. If the equation of state is v = \frac{RT}{P} + RTB(T), the \frac{RT}{P} and RTB contributions cancel in this combination, and only a term involving \frac{dB}{dT} remains, namely -RT^2\frac{dB}{dT}.

For the equation P(v-b) = RT, the internal energy is a function of temperature alone, because \left(\frac{\partial u}{\partial v}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_v - P = 0. Since h = u + Pv = h_0(T) + bP, the specific heat c_P = \frac{dh_0}{dT} depends on temperature only, and it does not depend on pressure or specific volume.

For an equation of state of the form P(v - C_1) = RT - \frac{C_2}{v}, the Maxwell relation \left(\frac{\partial s}{\partial v}\right)_T = \left(\frac{\partial P}{\partial T}\right)_v is the most efficient tool. Differentiating P with respect to T at constant v removes the C_2 term, which does not depend on T, and leaves \frac{R}{v - C_1}. The volume offset C_1 therefore appears in the result, while the C_2/v term does not.

Polytropic process of ideal gas

A polytropic process is one in which Pv^n = constant, where n is the polytropic index. Several familiar processes are special cases: n = 0 is isobaric, n = 1 is isothermal for an ideal gas, n = \gamma is reversible adiabatic (isentropic), and n \to \infty is isochoric. Here \gamma = c_P/c_V is the ratio of specific heats.

A relation of the form PV^{\gamma} = constant, with \gamma constant, describes the reversible adiabatic path of an ideal gas. Because such a path is quasi-static, a process that follows it is reversible, and a substance whose behaviour does not match the ideal-gas relation, such as a nearly incompressible liquid, cannot follow it.

For n \neq 1 the boundary work of a polytropic process is W_{12} = \frac{P_2V_2 - P_1V_1}{1-n}. For an ideal gas this becomes W_{12} = \frac{mR(T_2 - T_1)}{1-n}. The heat transfer is found from the polytropic specific heat c_n = c_v\frac{n - \gamma}{n - 1}, so that Q = m\,c_n\,(T_2 - T_1). A quick example: with \gamma = 1.4, n = 1.3, c_v = 0.718 kJ/(kg·K), m = 1 kg and a temperature rise of 100 K, c_n = -0.239 kJ/(kg·K) and Q = -23.9 kJ, so heat is rejected even though the gas is compressed.