| 3E-2 : | Hypothetical Process Paths and the Latent Heat of Vaporization | 4 pts |
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| Use the hypothetical process path shown here to help you determine the change in enthalpy in Joules for 20.0 g of heptane (C7H16) 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 path. 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. |  |
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| Read : | Step 1-2 is a bit tricky. We can use the Antoine Eqn with the Clausius-Clapeyron Eqn to estimate ΔHvap. Step 2-3 is straightforward because the problem instructs us to use an average Cp value. The only difficulty will be that Cp values may not be available at the temperatures of interest. Step 3-4 is cake because we were instructed to assume the heptane gas is ideal. As a result, enthalpy is not a function of pressure and ΔH3-4 = 0. | ||||||||
| Given: | m = | 20 |
g | Find: | ΔH1-2 | ??? |
J | ||
| T1 = | 300 |
K | ΔH2-3 | ??? |
J | ||||
| x1 = | 0.0 |
(sat'd liq) | ΔH3-4 | ??? |
J | ||||
| T2 = | 370 |
K | ΔH1-4 | ??? |
J | ||||
| P2 = | 58.7 |
kPa | |||||||
| 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 heat of vaporizatioon is constant over the temperature range of interest. | |||||||||
| - The superheated vapor also behaves as an ideal gas. | |||||||||
| - The heat capacity of the vapor is nearly constant over the temperature range of interest so that using the average value is a reasonable approximation. | |||||||||
| Solution : | First we can observe that : | ΔH1-2 = Latent heat of vaporization at 300 K | |||||||
| We can estimate the heat of vaporization using the Clausius -Clapeyron Equation. |
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Eqn 1 | |||||||
| If we plot Ln P* vs. 1/T(K), the slope is - ΔHvap/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. | |||||||||
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Eqn 2 |
Antoine Equation: | Log10(P*) = A - (B / (T + C)) | Eqn 3 | |||||
| P is in bar | T is in Kelvin |
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| The Antoine constants from the NIST WebBook are: | A = | 4.02832 | |||||||
| B = | 1268.636 | ||||||||
| C = | -56.199 | ||||||||
| From the Antoine Eqn: | Ta = | 299.5 |
K | Pa = | 6.52 |
kPa | |||
| Tb = | 300.5 |
K | Pb = | 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 | ΔHvap = | 36773 |
J/mol | ||||
 |
Eqn 4 |
MW = |
100.20 |
g/mol | n |
0.1996 |
mol | ||
| ΔH1-2 = | 7,340 |
J | |||||||
| Next, let's consider the enthalpy change from states 2 to 3, saturated vapor to superheated vapor: | |||||||||
| Because we assumed a constant heat capacity, we can evaluate ΔH using: | |||||||||
 |
Eqn 5 | ||||||||
| The heat capacities are tabulated in the NIST WebBook, under the Name Search option. Interpolate to estimate Cp at both T1 and T2. Then, average these two values of Cp to obtain the average heat capacity. This is equivalent to determining a linear equation between T1 and TT2 and integrating. | |||||||||
| Gas phase heat capacity data from the NIST WebBook: | T (K) |
Cp,gas (J/mol*K) |
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300 |
165.98 |
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400 |
210.66 |
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500 |
252.09 |
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| Solve: | |||||||||
| Cp(T1) = | 166.0 |
J/mole-K | There are many different ways to estimate Cp(T1) andCp(T2). | ||||||
| Cp(T2) = | 197.3 |
J/mole-K | |||||||
| Cp, avg = | 181.6 |
J/mole-K | ΔH(2-3) = | 12,713 | J/mol | ||||
| Now, just multiply by the number of moles, n, to get ΔH2-3 : | ΔH2-3 = | 2,538 |
J | ||||||
| Last, we need to determine the enthalpy change from states 3 to 4, in which the pressure of the superheated vapor is reduced. | |||||||||
| 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 : | |||||||||
| ΔH3-4 = | 0 |
J | |||||||
| Finally: | ΔH1-4 =ΔH1-2 +ΔH2-3 +ΔH3-4 = | 9,878 |
J | ||||||
Example Problem with Complete Solution
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