Example Problem with Complete Solution

3E-1 : Determination of the Vapor Pressure of Water
4 pts
The normal boiling point of water is 100°C and the latent heat of vaporization at this temperature is 2256.4 kJ/kg.  Estimate the vapor pressure of water at 85°C in kPa.  MWH2O = 18.016 g/mole, R = 8.314 J/mol-K.
                   
Read : The keys here are to know that the normal boiling point is the boiling point at 1atm and that the Clausius-Clapeyron Equation provides a relationship between the rate at which vapor pressure changes and the latent heat of vaporization.  Knowing that P*(100°C) = 101.325 kPa and the latent heat of vaporization at this temperature allows us to evaluate both the slope and the intercept in the Clausius-Clapeyron Equation and then use the result to estimate the vapor pressure at any other temperature.  We should keep in mind that this estimate is only reasonably accurate at temperatures close to the one known value, 100°C in this case.
Given : T1
100
 °C T2
85
 °C
373.15
 K
358.15
 K
P1*
101.325
 kPa ΔHvap
2256.4
 kJ/kg
MW
18.016
 g/mole R
8.314
 J/mole-K
Find : P2*
???
 kPa
Assumptions: - The Clausius-Clapeyron Equation applies.  This requires the following assumptions.
- 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.
Solution : We can estimate the heat of vaporization using the Clausius-Clapeyron Equation.
 
Equation. Eqn 1
If we plot Ln P* vs. 1/T(K), the slope is -ΔHvap/R.  Don't forget to use T in Kelvins in Eqn 1.
So, the next thing we need to do is use the given value of the latent heat to estimate this slope.
Equation.
Eqn 2
 ΔHvap
40651
 J/mole
Slope
-4889.5
 K
Next, we can use the one known value of the vapor pressure (at 100°C) to evaluate the constant (C) in the Clausius-Clapeyron Equation.
 
Equation.
Eqn 3
 C
17.722
Evaluating C in this manner has a catch.  This value of C only applies as long as the same units of pressure are used in Eqn 1.  Since we used P1* in kPa, we must always use P in kPa whenever we use this value of C.
Now, we can use the values of the slope and intercept that we have determined to subsititute back into Eqn 1 to estimate the vapor pressure of water at a temperature other than 100°C, in this case 85°C.
P2*
58.53
 kPa
   
Verify: The assumptions made in this problem solution cannot be verified.
 
Ideal Gas EOS : Equation. Eqn 4
Solve for molar volume : Equation. Eqn 5
Plug in values based upon the results we obtained above :
V1
3.06E-02
 m3/mol V2
5.09E-02
 m3/mol
30.6
 L/mol
50.9
 L/mol
Because the molar volume of the saturated vapor at both (100°C, 101.325 kPa) and (85°C, 58.5 kPa) is greater than 20 L/mole, it is accurate to treat the saturated vapors as ideal gases.
The 2nd assumption required to use the Clausius-Clapeyron Equation cannot be verified with the information provided in the problem statement.  However, based on data available in the Steam Tables, this assumption is also valid under the conditions in this problem.  So it is valid to use the Clausius-Clapeyron Equation.
The Steam Tables also tell us that: P2*
57.9
 kPa
Our answer based on the Clausius-Clapeyron Equation is not accurate in the second digit.  This is caused by the fact that the latent heat of vaporization is not constant over the wide range of temperatures in this problem (100°C to 85°C).  The latent heat of vaporization of water at 85°C is 2295.3 kJ/kg.  This is a difference of almost than 2%.  This is the principle source of error in using the Clausius-Clapeyron Equation for this problem.
Answers : P2*
58.5
 kPa
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