Ideal gases, real gases and specific heats

Steam tables are indispensable for water, but many of the fluids at Northgate are gases far from condensing: combustion air drawn into the boiler burners, the flue gas leaving the stacks, natural gas arriving through the utility's supply line, and the compressed air in the instrument-air receiver. For these, a single simple equation relates pressure, volume and temperature with good accuracy, and internal energy and enthalpy depend on temperature alone. This lesson develops the ideal-gas model, shows when it fails and how to correct it, and introduces the specific heats that link energy to temperature for gases, liquids and solids.

The ideal-gas equation of state

An equation of state relates the pressure, temperature and specific volume of a substance. The simplest and best known is the ideal-gas equation, which grew out of the experiments of Boyle, Charles and Gay-Lussac: at low density, the pressure of a gas is proportional to its absolute temperature and inversely proportional to its specific volume.

Here \(R\) is the gas constant of the particular gas, \(m\) its mass, \(N\) the number of moles, and \(R_u = 8.314\ \text{kJ/(kmol·K)}\) the universal gas constant, the same for all gases. The two are related through the molar mass \(M\): \(R = R_u/M\). For air, \(M = 28.97\ \text{kg/kmol}\) and \(R = 0.287\ \text{kJ/(kg·K)}\); for methane, the main component of natural gas, \(M = 16.04\) and \(R = 0.5183\); for steam, \(M = 18.02\) and \(R = 0.4615\). Pressure must be absolute and temperature in kelvins. With pressure in kPa and volume in m³, \(PV\) comes out in kJ, which matches \(R\) in kJ/(kg·K).

A convenient consequence: for a fixed mass of ideal gas, \(P_1V_1/T_1 = P_2V_2/T_2\). This combined gas law lets us find a third property after a change in two others, without knowing the gas constant at all.

When is a gas ideal?

The ideal-gas model assumes the molecules occupy negligible volume and exert no forces on one another except during collisions. Both assumptions hold well when the gas is at low density, that is, at low pressure or high temperature compared with its critical values. Air at room conditions, at a pressure far below its critical pressure of about 3.8 MPa and a temperature far above its critical temperature of about 133 K, is ideal to within a fraction of a percent. Combustion gases in the boiler stack, at high temperature and atmospheric pressure, are also nearly ideal.

Steam is a different matter. Close to the saturated vapour line, or at high pressures, the ideal-gas equation can be wrong by many percent; at 4 MPa and 400 °C the error in specific volume is about 6 percent, and near the critical point it can exceed 100 percent. At low pressures, below about 10 kPa, steam is ideal to better than 0.1 percent at any temperature, which is why water vapour in the atmosphere can be treated as an ideal gas in air-conditioning calculations. For steam in power plants and district systems, always use the steam tables or software.

The compressibility factor

The deviation from ideal-gas behaviour is measured by the compressibility factor \(Z\), defined as

For an ideal gas \(Z = 1\). For real gases \(Z\) can be smaller than 1, when attractive forces between molecules pull them closer than in an ideal gas, or larger than 1, at very high pressures where the molecules' own volume dominates. Remarkably, when pressure and temperature are expressed as fractions of their critical values, as reduced pressure \(P_R = P/P_{cr}\) and reduced temperature \(T_R = T/T_{cr}\), the compressibility factor of all gases is approximately the same function of \(P_R\) and \(T_R\). This principle of corresponding states lets one generalized compressibility chart serve for any gas.

The chart gives three useful rules. At very low reduced pressure, \(P_R \ll 1\), gases behave ideally whatever the temperature. At high reduced temperature, \(T_R > 2\), the ideal-gas model is good up to fairly high pressures, \(P_R\) of about 4. The largest deviations occur near the critical point, \(P_R \approx 1\) and \(T_R \approx 1\). For methane at 15 °C, \(T_R = 288/190.6 = 1.51\); in the utility's transmission main at 5 MPa, \(P_R = 1.09\), and the generalized chart gives \(Z \approx 0.89\), an 11 percent deviation that the gas company must account for when it bills by volume.

Other equations of state

Many equations of state improve on the ideal-gas law. The first and simplest was proposed by van der Waals in 1873:

The term \(a/v^2\) accounts for intermolecular attraction, and \(b\) for the volume occupied by the molecules. The constants follow from the requirement that the critical isotherm has a horizontal inflection at the critical point. Van der Waals' equation captures the qualitative behaviour of real gases, including condensation, but it is not very accurate. Later equations, such as Redlich-Kwong, Peng-Robinson and the many-constant Benedict-Webb-Rubin equation, are far better and are used in process simulation software and in the natural-gas industry.

The program below compares the ideal-gas and van der Waals specific volumes for methane at 15 °C over a range of pipeline pressures. At 100 kPa they agree within 0.2 percent; at 5 MPa van der Waals gives \(Z = 0.887\), close to the chart value; at 10 MPa it gives 0.786, rather lower than the chart's 0.83, illustrating that the simple equation exaggerates the deviation at high pressure. It also repeats the air-receiver calculation.

# Ideal gas versus van der Waals for natural gas (taken as methane) in the plant's supply line
Ru = 8.314                          # kJ/(kmol.K)
M = 16.04                           # kg/kmol, methane
R = Ru / M                          # kJ/(kg.K)
Tc, Pc = 190.6, 4599.0              # K, kPa
a = 27 * R**2 * Tc**2 / (64 * Pc)   # van der Waals constants on a mass basis
b = R * Tc / (8 * Pc)

def v_vdw(P, T):
    """Solve P = RT/(v - b) - a/v^2 for the gas-like root by Newton's method."""
    v = R * T / P
    for _ in range(50):
        f = R * T / (v - b) - a / v**2 - P
        df = -R * T / (v - b)**2 + 2 * a / v**3
        v -= f / df
    return v

T = 288.15                          # K (15 C)
print(f"R = {R:.4f} kJ/kg.K, a = {a:.4f} kPa.m6/kg2, b = {b:.6f} m3/kg")
print(f"{'P kPa':>7s} {'v ideal':>9s} {'v vdW':>9s} {'Z = Pv/RT':>10s}")
for P in (100, 400, 2000, 5000, 10000):
    v = v_vdw(P, T)
    print(f"{P:7d} {R * T / P:9.5f} {v:9.5f} {P * v / (R * T):10.4f}")

# Mass of air in the instrument-air receiver: 2.0 m3 at 800 kPa gauge, 25 C
Rair, Patm = 0.287, 99.0
P, V, Tair = 800 + Patm, 2.0, 298.15
m = P * V / (Rair * Tair)
print(f"\nair receiver: m = {m:.2f} kg at {P:.0f} kPa abs")
# Pressure after the room cools the air to 10 C (rigid tank, same mass)
print(f"cooled to 10 C: P = {m * Rair * 283.15 / V:.0f} kPa abs")

R = 0.5183 kJ/kg.K, a = 0.8953 kPa.m6/kg2, b = 0.002685 m3/kg
  P kPa   v ideal     v vdW  Z = Pv/RT
    100   1.49357   1.49025     0.9978
    400   0.37339   0.37007     0.9911
   2000   0.07468   0.07132     0.9551
   5000   0.02987   0.02648     0.8865
  10000   0.01494   0.01174     0.7859

air receiver: m = 21.01 kg at 899 kPa abs
cooled to 10 C: P = 854 kPa abs

Specific heats

The specific heat is the energy needed to raise the temperature of a unit mass of a substance by one degree. Its value depends on how the heating is done. The specific heat at constant volume, \(c_v\), is the energy needed per degree when the volume is held fixed; the specific heat at constant pressure, \(c_p\), is the energy needed when the pressure is held fixed, so that the substance also does boundary work as it expands. Precisely, they are defined as partial derivatives of properties, which makes them properties too:

Because \(c_p\) includes the energy that goes into boundary work, it is always at least as large as \(c_v\). Both are in kJ/(kg·K), which is numerically the same as kJ/(kg·°C) because a kelvin and a Celsius degree are the same size.

Internal energy and enthalpy of ideal gases

Joule showed experimentally, and statistical mechanics confirms, that the internal energy of an ideal gas depends on temperature only. Since \(h = u + Pv = u + RT\), enthalpy also depends on temperature only. The partial derivatives above then become ordinary derivatives, and changes in \(u\) and \(h\) follow from integrating the specific heats over temperature. Over moderate temperature ranges, constant average values give good accuracy:

These relations hold for any process of an ideal gas, not only constant-volume or constant-pressure ones, because \(u\) and \(h\) are functions of temperature alone. That is a powerful simplification. Differentiating \(h = u + RT\) gives the relation between the two specific heats, and their ratio defines the specific heat ratio \(k\):

Monatomic gases such as helium and argon have nearly constant specific heats with \(k = 5/3\). For diatomic and polyatomic gases the specific heats rise with temperature, as molecular rotation and vibration absorb more energy. For air, \(c_p\) rises from 1.005 kJ/(kg·K) at 300 K to about 1.14 at 1000 K. When temperature changes are large, as in the boiler furnace or a gas turbine, use variable specific heats from ideal-gas tables, which list \(h\) and \(u\) against temperature; for the modest changes in compressors and heat exchangers at Northgate, constant values at the average temperature are accurate to a percent or so.

Liquids and solids

Liquids and solids are nearly incompressible: their specific volumes change very little with pressure. For such substances \(c_p\) and \(c_v\) are essentially equal, and a single specific heat \(c\) is used. Internal energy depends almost entirely on temperature, \(\Delta u \approx c_{\text{avg}}\Delta T\). Enthalpy picks up a small pressure term:

For liquid water, \(c \approx 4.18\ \text{kJ/(kg·K)}\), and the pressure term is small unless the pressure change is large: raising water by 4 MPa in the feedwater pump adds \(v\Delta P \approx 0.00105(4000) = 4.2\ \text{kJ/kg}\), the same as a 1 °C temperature rise. That small number is the work needed per kilogram to pump water into a boiler, and Lesson 12 will show why it is so much smaller than the work a turbine extracts from the steam. For steel, \(c \approx 0.46\ \text{kJ/(kg·K)}\), which matters when estimating how long a cold boiler takes to warm up.