3E3 :  Hypothetical Process Paths and the Latent Heat of Vaporization  4 pts 

Use the hypothetical process path (HPP) shown here to help you determine DH in Joules for 32.5 g of heptane (C_{7}H_{16}) as it changes from a saturated liquid at 300 K to a temperature of 370 K and a pressure of 58.7 kPa.  


Calculate the ΔH for each step in the HPP. Do not use tables of thermodynamic properties, except to check your answers. Instead, use the Antoine Equation to estimate the heat of vaporization of heptane at 300 K.  
Use the average heat capacity of heptane gas over the temperature range of interest. Assume heptane gas is an ideal gas at the relevant temperatures and pressures.  
Read :  Step
12 is a bit tricky. We can use the Antoine
Equation with the ClausiusClapeyron
Equation to estimate DH_{vap}. Step 23 is straightforward because the problem instructs us to use an average C_{p} value. The only difficulty will be that C_{p} values may not be available at the temperatures of interest. Step 34 is cake because we were instructed to assume the heptane gas is ideal. As a result, enthalpy is not a function of pressure and DH_{34} = 0. 

Diagram:  The hypothetical process path diagram in the problem statement is adequate.  
Given:  m  32.5  g  Find:  DH_{12}  ???  J  
T_{1} = T_{2}  300  K  DH_{23}  ???  J  
x_{1}  0  kg vap/kg  (sat'd liq)  DH_{34}  ???  J  
T_{3} = T_{4}  370  K  DH_{14}  ???  J  
P_{4}  58.7  kPa  
Assumptions:  1  ClausiusClapeyron applies:  
1a The saturated vapor is an ideal gas  
1b The molar volume of the saturated vapor is much, much greater than the molar volume of the saturated liquid.  
1c The latent heat of vaporization is constant over the temperature range of interest.  
2  The superheated vapor also behaves as an ideal gas.  
3  The heat capacity of the superheated vapor is nearly linear with respect to temperature over the temperature range of interest so that using the average value is a reasonable approximation.  
Equations / Data / Solve:  
First we can observe that:  DH_{12} = Latent heat of vaporization at 300 K  
We can estimate the latent heat of vaporization using the Clausius Clapeyron Equation. 

Eqn 1  
If we plot Ln P* vs. 1/T(K), the slope is  DH_{vap}/R.  
We can calculate the vapor pressures at two different temperatures using the Antoine Equation. Use temperatures near the temperature of interest, 300 K. Use the two points to estimate the slope over this small range of temperatures.  

Eqn 2  
Antoine Equation:  Log_{10}(P*) = A  (B / (T + C))  Eqn 3  
P is in bar  T is in Kelvin  
The Antoine constants from the NIST WebBook are:  A =  4.02832  
B =  1268.636  
C =  56.199  
From the Antoine Equation:  
T_{1} = T_{2}  300  K  P_{1} = P_{2} = P  6.68  kPa  
T_{a}  299.5  K  P_{a} =  6.52  kPa  
T_{b}  300.5  K  P_{b} =  6.85  kPa  
Slope  4423.1  K  
Next we use this slope with Eqn 1 to determine the latent heat of vaporization at 300 K :  
R =  8.314  J/mol K  DH_{vap} =  36773  J/mol  

Eqn 4  MW  100.20  g/mol  
n  0.3244  mol  
DH_{(12)} =  11,928  J  
Next, let's consider the enthalpy change from states 2 to 3, saturated vapor to superheated vapor.  
The enthalpy change associated with a temperature change for an ideal gas can be determined from :  

Eqn 5  
Because we assumed a constant heat capacity, Eqn 4 simplifies to:  

Eqn 6  
The heat capacities are tabulated in the NIST WebBook, under the Name Search option. Interpolate to estimate C_{p} at both T_{1} and T_{2}. Then, average these two values of C_{p} to obtain the average heat capacity. This is equivalent to determining a linear equation between T_{1} and T_{2} and integrating.  
Gas phase heat capacity data from the NIST WebBook:  Temperature (K)  Cp,gas (J/mol*K)  
300  165.98  
400  210.66  
500  252.09  
There are many different ways to estimate C_{p}(T_{1}) and C_{p}(T_{2}).  
C_{p}(T_{1}) =  166.0  J/moleK  
C_{p}(T_{2}) =  197.3  J/moleK  
C_{p, avg} =  181.6  J/moleK  DH_{(23)} =  12,713  J/mol  
Now, just multiply by the number of moles, n, to get DH_{23} :  DH_{(23)} =  4,124  J  
Last, we need to determine the enthalpy change from states 3 to 4, in which the pressure of the superheated vapor is increased.  
Recall the assumption that the vapor behaves as an ideal gas. Because enthalpy is only a function of T for ideal gases, and since T_{3} = T_{4} :  
DH_{(34)} =  0  J  
Finally:  DH_{14} = DH_{12} + DH_{23} + DH_{34} =  16,051  J  
Verify:  The problem statement instructed us to make all of the assumptions that we used.  
Nonetheless, we will verify the assumptions as well as we can from the given information.  
1a  Is the saturated vapor is an ideal gas ? 

T_{1}  300  K  
Use the Antoine Equation to determine P_{1} :  P_{1}  6.68  kPa  
V_{1}  373.4  L/mol  
Since V_{1} > 21 L/mole this ideal gas assumption is valid.  
1b  Is the molar volume of the saturated vapor is much, much greater than the molar volume of the saturated liquid.  
1c  Is the latent heat of vaporization is constant over the temperature range of interest.  
We cannot assess the validity of assumptios 1b and 1c from the data given in the problem.  
2  Does the superheated vapor also behave as an ideal gas. 


T_{3} = T_{4}  370  K  
P_{4}  58.70  kPa  V_{4}  52.4  L/mol  
Since V_{4} > 21 L/mole this ideal gas assumption is valid.  
3  Is the heat capacity of the superheated vapor is nearly constant over the temperature range of interest ?  
We cannot verify this assumption with the data provided in the problem statement.  
We have no evidence that any of the assumptions are invalid.  
Answers :  DH_{12}  11,900  J  ( All rounded to 3 significant digits)  
DH_{23}  4,120  J  
DH_{34}  0  J  
DH_{15}  16,100  J 