# Example Problem with Complete Solution

3E-1 : Hypothetical Process Paths and the Latent Heat of Vaporization 7 pts
Use the hypothetical process path (HPP) shown below to help you determine the DH 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 DH for each step in the HPP. 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.

Read : Step 1-2 is straightforward because we will assume that the liquid propane is incompressible.
We can use the
Antoine Equation with the Clausius-Clapeyron Equation to estimate DHvap for step 2-3.
Step 3-4 is easy 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.
Diagram: The diagram in the problem statement is adequate.
Given: P1 300 kPa Find: DH1-2 ??? J/mol
T1 250 K DH2-3 ??? J/mol
T5 300 K DH3-4 ??? J/mol
P5 100 kPa DH4-5 ??? J/mol
Vliq 7.8914E-05 m3/mole DH1-5 ??? J/mol
CoP 69.0 J/mole-K
Assumptions: 1 - 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 latent heat of vaporization is constant over the temperature range of interest.
2 - The superheated vapor also behaves as an ideal gas.
3 - Liquid propane is incompressible.
Equations / Data / Solve:
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 : 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.
DH12 -5.768 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. 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. Eqn 5
From the Antoine Equation:
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 DHvap 18197 J/mole
DH23 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 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 :
DH34 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: Eqn 6
Plugging numbers into Eqn 6 yields : DH45 = 3,450 J/mole
Finally, put them all together:   DH15 = DH12 + DH23 + DH34 + DH45 = 21,641 J/mole
Notice that DH12 is very small compared to ΔH23 and ΔH45. In fact ΔH12 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.
Verify: 1 - We can test the validity of the ideal gas assumption for state 3 as follows. V3 9.160 L/mol
Because V3 < 20 L/mole, the Clausius-Clapeyron Equation is not very accurate. This issue makes the results from this analysis somewhat unreliable.
It is not as easy to test the 2nd assumption that underpins the Clausius-Clapeyron Equation.
We can use the NIST Webbook to determine the molar volume of saturated liquid and saturated vapor at 250 K.
Vsat vap 8.9258 L/mol
Vsat liq 0.078977 L/mol Vsat vap / Vsat liq = 113.02
Since Vsat vap is more than 100 times greater than Vsat liq this assumption underpinning the use of the Clausius-Clapeyron Equation is valid.
Because we considered a very narrow temperature range, just 0.2°C, the last assumption underpinning the use of the Clausius-Clapeyron Equation is almost certainly valid.
2 - Is the superheated vapor be accurately treated as an ideal gas? V3 20.785 L/mol
Because V4 > 20 L/mole, the Clausius-Clapeyron Equation can be applied.
3 - Since ΔH12 is negligible, this assumption is not very important.
Nonetheless, we can use the NIST Webbook to determine the molar volume of liquid at P1 = 100 kPa and at P2 = 226.9 kPa at 250 K. and see if the molar volume changes significantly.
V1 0.0625 L/mol 10.887059 V2 0.0693 L/mol
We find that V2 differs from V1 by about 11%. So it is not very accurate to treat liquid propane as an incompressible liquid under these conditions.
This may be ok in this problem since ~ΔH12 is so small that an 11% error in its value will still not matter. 