Example Problem with Complete Solution

3D-1 : Hypothetical Process Paths and the Latent Heat of Vaporization 7 pts
Use the hypothetical process path shown below to help you determine the change in enthalpy in J/mole for propane (C3H8) as it changes from a subcooled liquid at P1 = 300 kPa and T1 = 250 K to a superheated vapor at P5 = 100 kPa and T5 = 300 K.  Calculate the molar ΔH for each step in the path.  Assume the propane vapor behaves as an ideal gas and a constant heat capacity of 69.0 J/mole-K.  Do not use tables of thermodynamic properties, except to check your answers. Use the Antoine and Clausius-Clapeyron Equations to estimate the heat of vaporization of propane at T1.  Note: The molar volume of saturated liquid propane at 250 K is 7.8914 x 10-5 m3/mole.
               
Figure 1
               
Read : Step 1-2 is straightforward because we will assume that the liquid propane is incompressible.  We can use the Antoine Eqn with the Clausius-Clapeyron Eqn to estimate DHvap for step 2-3.   Step 3-4 is cake because we were instructed to assume the propane is an ideal gas and the enthalpy of an ideal gas is not a function of pressure.  Step 4-5 is straightforward because the problem instructs us to use a constant Cp value. 
Given: Find:
P1 300.0 kPa ΔH1-2 ??? J
T1 250 K ΔH2-3 ??? J
T5 300 K ΔH3-4 ??? J
P5 100.0 kPa ΔH4-5 ??? J
Vliq 7.8914E-05 m3/mole ΔH1-5 ??? J
CoP 69.0 J/mole-K
Assumptions : - Clausius-Clapeyron applies:
- The saturated vapor is an ideal gas
- The molar volume of the saturated vapor is much much greater than the molar volume of the saturated liquid.
- The superheated vapor also behaves as an ideal gas.
- Liquid propane is incompressible.
Equations : Step 1-2 involves a change in pressure on an incompressible liquid at constant temperature.
Since neither the internal energy nor the molar volume of an incompressible liquid are functions of pressure :
Equation 1 Eqn 1
We can use the Antoine Equation to determine the vapor or saturation pressure of propane at T1.
Log10(P*) = A - (B / (T + C)) Eqn 2
P is in bar T is in Kelvin
The Antoine constants from the NIST WebBook are: A = 4.53678
B = 1149.36
C = 24.906
P2 = P*(T1) Eqn 3 P2 226.9 kPa
Now, we can plug numbers into Eqn1,
but be careful with the units.
ΔH12 -0.00577 J/mole
Next, we can observe that DH23 = Latent heat of vaporization at 250 K.
We can estimate the heat of vaporization using the Clausius -Clapeyron Equation.
Equation 3 Eqn 4
If we plot Ln P* vs. 1/T(K), the slope is - DHvap/R.
We can calculate the vapor pressures at two different temperatures using the Antoine equation.  Use temperatures near the temperature of interest, 250 K.  Use the two points to estimate the slope over this small range of temperatures.
Equation 4 Eqn 5
From the Antoine Eqn:
Ta =   249.9 K Pa =  226.12 kPa
Tb =   250.1 K Pb =  227.71 kPa
Slope = -2188.7 K
Next we use this slope with Eqn 4 to determine the latent heat of vaporization at
250 K :
R = 8.314 J/mol K ΔHvap 18197 J/mole
ΔH23 18,197 J/mole
Next, we need to determine the enthalpy change from state 3 to 4, in which the pressure of the saturated vapor vapor is reduced.  This causes the vapor to become a superheated vapor.
Recall the assumption that the vapor behaves as an ideal gas.  Because enthalpy is only a function of T for ideal gases, and since T3 = T4 :
ΔH34 0 J/mole
Next, let's consider the enthalpy change from state 4 to 5.
Because we assumed the vapor phase is an ideal gas with constant CP, we can evaluate DH using:
Equation 5 Eqn 6
Plugging numbers into Eqn 6 yields : ΔH45 = 3,450 J/mole
Finally, put them all together:  
ΔH15 = ΔH12 + ΔH23 + ΔH34 + ΔH45 = 21,647 J/mole
Notice that DH12 is so small that it is negligible.
This shows why it is often acceptable to approximate the enthalpy of a subcooled liquid using the enthalpy of the saturated liquid at the same TEMPERATURE.  It is NOT accurate to approximate the enthalpy of a subcooled liquid using the enthalpy of the saturated liquid at the same PRESSURE.