If a Carnot engine is operated between a high‐temperature source at the temperature of boiling water TH and a low‐temperature sink at the temperature of melting ice TL, and the amount of heat received Qsteam and rejected Qice were measurable accurately, the ratio Qsteam/Qice would be equal to 1.3662. Since Qsteam/Qice = Tsteam/Tice,
(1.90)
If the temperature range Tsteam − Tice is arbitrarily divided into 100 equal divisions (call it 100 °C), a second equation can be obtained
(1.91)
The simultaneous solution of Eqs. (1.90) and (1.91) yields the temperature of boiling water and melting ice on a thermodynamic scale:
If the temperature range Tsteam − Tice is divided into 180 equal divisions (call it degrees Fahrenheit, F), we get
(1.92)
The determination of a thermodynamic scale using reversible Carnot engine as described here is practically impossible. To determine where the number 1.3662 comes from, we start with the equation of state for a perfect gas at constant volume and constant pressure, which yields
(1.93)
Next, constant‐volume and constant‐pressure thermometers are placed alternately in boiling water and melting ice, and the volume and pressure ratios of the gases in the devices (carbon dioxide CO2, hydrogen H2, and helium He) are measured at an initial pressure. If the procedure is repeated for different initial pressures and the results plotted versus pressure, straight lines are obtained, which converge at zero absolute pressure to a value of 1.3662, which must be the ratio of absolute temperature of boiling water and melting ice.
Currently, the international temperature scale (ITS‐90) is the standard used to represent the thermodynamic (absolute) temperature scale as closely as possible with improved accuracy over gas thermometry.
1.3.9 Third Law of Thermodynamics
This law, based on empirical evidence, postulates that absolute entropy of a pure crystalline substance in complete internal equilibrium is zero at temperature zero degree absolute. The third law allows the determination of absolute entropies from thermal data.
The entropy change of a gas on molar basis is
The entropy change for process 1–2 is then
(1.94)
If the specific heat Cp is assumed constant, Eq. (1.94) becomes
(1.95)
More accurate results can be obtained if the variability of specific heat with temperature is accounted for. Taking the absolute zero as the reference temperature,
(1.96)
The values of
(1.97)
Substituting Eq. (1.97) in Eq. (1.94), we obtain
(1.98)
As Eq. (1.98) shows, entropy changes with both temperature and pressure. When using Eq. (1.98) in chemical reactions, the pressure ratio in the last term is replaced by the mole concentration of each substance.
The
(1.99)
The coefficients of correlation Eq. (1.99) for six gases are given in Table 1.3.
Table 1.3 Coefficients of Eq. (1.99) for the calculation of
of combustion products, pref = 0.1 MPa Tref = 0 K.
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Coefficient | CO2 | CO | H2O | H2 | O2 | N2 |