3 Ways To Calculate Vapor Pressure

ΔHvap: The enthalpy of vaporization of the liquid. This can usually be found in a table at the back of chemistry textbooks. R: The real gas constant, or 8. 314 J/(K × Mol). T1: The temperature at which the vapor pressure is known (or the starting temperature. ) T2: The temperature at which the vapor pressure is to be found (or the final temperature. ) P1 and P2: The vapor pressures at the temperatures T1 and T2, respectively.

For example, let’s say that we’re told that we have a container full of liquid at 295 K whose vapor pressure is 1 atmosphere (atm). Our question is: What is the vapor pressure at 393 K? We have two temperature values and a pressure, so we can solve for the other pressure value with the Clausius-Clapeyron equation. Plugging in our variables, we get ln(1/P2) = (ΔHvap/R)((1/393) - (1/295)). Note that, for Clausius-Clapeyron equations, you must always use Kelvin temperature values. You can use any pressure values as long as they are the same for both P1 and P2.

In our example, let’s say that our liquid is pure liquid water. If we look in a table of ΔHvap values, we can find that the ΔHvap is roughly 40. 65 kJ/mol. Since our H value uses joules, rather than kilojoules, we can convert this to 40,650 J/mol. Plugging our constants in to our equation, we get ln(1/P2) = (40,650/8. 314)((1/393) - (1/295)).

The only difficult part of solving our equation (ln(1/P2) = (40,650/8. 314)((1/393) - (1/295))) is dealing with the natural log (ln). To cancel out a natural log, simply use both sides of the equation as the exponent for the mathematical constant e. In other words, ln(x) = 2 → eln(x) = e2 → x = e2. Now, let’s solve our equation: ln(1/P2) = (40,650/8. 314)((1/393) - (1/295)) ln(1/P2) = (4,889. 34)(-0. 00084) (1/P2) = e(-4. 107) 1/P2 = 0. 0165 P2 = 0. 0165-1 = 60. 76 atm. This makes sense — in a sealed container, increasing the temperature by almost 100 degrees (to almost 20 degrees over the boiling point of water) will create lots of vapor, increasing the pressure greatly

Psolution: The vapor pressure of the entire solution (all of the component parts combined) Psolvent: The vapor pressure of the solvent Xsolvent: The mole fraction of the solvent. Don’t worry if you don’t know terms like “mole fraction” — we’ll explain these in the next few steps.

Let’s work through a simple example in this section to illustrate the concepts we’re discussing. For our example, let’s say that we want to find the vapor pressure of simple syrup. Traditionally, simple syrup is one part sugar dissolved in one part water, so we’ll say that sugar is our solute and water is our solvent. Note that the chemical formula for sucrose (table sugar) is C12H22O11. This will be important soon.

In our example, let’s say that the simple syrup’s current temperature is 298 K ( about 25 C).

The Clausius-Clapeyron can help here — use the reference vapor pressure and 298 K (25 C) for P1 and T1 respectively. In our example, our mixture is at 25 C, so we can use our easy reference tables. We find that water at 25 C has a vapor pressure of 23. 8 mm HG [7] X Research source

Let’s say that our recipe for simple syrup uses 1 liter (L) of water and 1 liter of sucrose (sugar. ) In this case, we’ll need to find the number of moles in each. To do this, we’ll find the mass of each, then use the substance’s molar masses to convert to moles. Mass (1 L of water): 1,000 grams (g) Mass (1 L of raw sugar): Approx. 1,056. 7 g[9] X Research source Moles (water): 1,000 grams × 1 mol/18. 015 g = 55. 51 moles Moles (sucrose): 1,056. 7 grams × 1 mol/342. 2965 g = 3. 08 moles (note that you can find sucrose’s molar mass from its chemical formula, C12H22O11. ) Total moles: 55. 51 + 3. 08 = 58. 59 moles Mole fraction of water: 55. 51/58. 59 = 0. 947

Substituting our values, we get: Psolution = (23. 8 mm Hg)(0. 947) Psolution = 22. 54 mm Hg. This makes sense — in mole terms, there’s only a little sugar dissolved in a lot of water (even though in real-world terms the two ingredients have the same volume), so the vapor pressure will only decrease slightly.

Temperature: 273. 15 K / 0 C / 32 F Pressure: 760 mm Hg / 1 atm / 101. 325 kilopascals

For instance, let’s say that we have an unknown liquid with a vapor pressure of 25 torr at 273 K and 150 torr at 325 K and we want to find this liquid’s enthalpy of vaporization (ΔHvap). We could solve like this: ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)) (ln(P1/P2))/((1/T2) - (1/T1)) = (ΔHvap/R) R × (ln(P1/P2))/((1/T2) - (1/T1)) = ΔHvap Now, we plug in our values: 8. 314 J/(K × Mol) × (-1. 79)/(-0. 00059) = ΔHvap 8. 314 J/(K × Mol) × 3,033. 90 = ΔHvap = 25,223. 83 J/mol

For example, let’s say that we have a solution made from two chemicals: benzene and toluene. The total volume of the solution is 120 milliliters (mL); 60 mL of benzene and 60 of toluene. The temperature of the solution is 25 C and the vapor pressures of each of these chemicals at 25 C is 95. 1 mm Hg for benzene 28. 4 mm Hg for toluene. Given these values, find the vapor pressure of the solution. We can do this as follows, using standard density, molar mass, and vapor pressure values for our two chemicals: Mass (benzene): 60 mL = . 060 L × 876. 50 kg/1,000 L = 0. 053 kg = 53 g Mass (toluene): . 060 L × 866. 90 kg/1,000 L = 0. 052 kg = 52 g Moles (benzene): 53 g × 1 mol/78. 11 g = 0. 679 mol Moles (toluene): 52 g × 1 mol/92. 14 g = 0. 564 mol Total moles: 0. 679 + 0. 564 = 1. 243 Mole fraction (benzene): 0. 679/1. 243 = 0. 546 Mole fraction (toluene): 0. 564/1. 243 = 0. 454 Solve: Psolution = PbenzeneXbenzene + PtolueneXtoluene Psolution = (95. 1 mm Hg)(0. 546) + (28. 4 mm Hg)(0. 454) Psolution = 51. 92 mm Hg + 12. 89 mm Hg = 64. 81 mm Hg