Example Problem with Complete Solution

3E-2 : Determination of the Vapor Pressure of Ammonia 4 pts
Use the Clausius-Clapeyron equation to estimate the vapor pressure, in kPa, of ammonia at -25oC. The normal boiling point of ammonia is -33.34oC and the latent heat of vaporization at this temperature is 1370 kJ/kg. ammonia SI with C
 
151.4713
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*(-33.34oC) = 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, -33.34oC in this case.
101.2561 151.4713
Diagram: A diagram is not needed in the solution of this problem. 1399.072
29.53419
Given: T1 -33.34 oC T2 -25 oC 1369.538
239.81 K 248.15 K 23.32359
P1* 101.325 kPa DHvap 1370 kJ/kg
MW 17.03 g/mole R 8.314 J/mole-K
Find: P2* ??? kPa
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.
Equations / Data / Solve:
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 - DHvap/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.
Eqn 2 DHvap 23331 J/mole
Slope -2806.242 K
Next, we can use the one known value of the vapor pressure (at -33.34oC) to evaluate the constant (C) in the Clausius-Clapeyron Equation.
Eqn 3 C 16.320
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 ammonia at a temperature other than -33.34oC, in this case -25oC.
P2* 150.1 kPa
Verify: Only the ideal gas assumption can be verified using the data in the problem statement.
Ideal Gas EOS :
Eqn 4
Solve for molar volume :
Eqn 5
Plug in values based upon the results we obtained above :
V1 1.97E-02 m3/mol V2 1.37E-02 m3/mol
19.7 L/mol 13.7 L/mol
Because the molar volume of the saturated vapor at both (-33.34oC, 101.325 kPa) and (-25oC, 150.1 kPa) is less than 20 L/mole, it is not accurate to treat the saturated vapors as ideal gases. This is a more serious problem at -25C and 150.1 kPa.
The 2nd and 3rd assumptions 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 Ammonia Tables, these two assumption are valid under the conditions in this problem.
The Ammonia Tables also tell us that: P2* 151.5 kPa
Our answer based on the Clausius-Clapeyron Equation is accurate to within about 1%. This is surprisingly good in light of the fact that the ideal gas assumption for the saturated vapor is not valid!
Answers: P2* 150.1 kPa