The vapor pressure is calculated by calculating some values such as the initial vapor pressure, the temperature at that vapor pressure, the enthalpy of vaporization, and the temperature at which you want to calculate the vapor pressure. The Clausius-Clapeyron equation is used to define the relationship between vapor pressure and temperature. The pressure that a vapor in thermodynamic equilibrium exerts on its condensed phases solid or liquid at a specific temperature within a closed system is known as vapor pressure.
What is Vapor Pressure?
Vapor is defined as a converted matter of solid or liquid into a gaseous state. Example: Heating of water at a certain temperature. Pressure is defined as force applied to a unit area. Pressure = force/ area newton/m2. When the water in a closed container is heated, the thermal energy increase the kinetic energy of molecules then matters in the liquid state are converted into a gaseous state. Molecules in this gaseous state apply pressure against the walls and lid of the container, this applied pressure by the vapor is called Vapor pressure.
Saturated vapor pressure
The point at which a number of molecules escaping the liquid or solid is equal to the number of molecules returning from vapor to liquid or solid.
- Case 1: In a closed container, evaporation rate and condensation rate are equal at saturated vapor pressure.
- Case 2: In an open container, vapor pressure rises as temperature increases until temperature reaches boiling point, which depends on atmospheric pressure.
So, at 1 atm of pressure, the saturated vapor pressure of water occurs at 100°C. At the liquids boiling point, vapor pressure is equal to the atmospheric pressure.
Vapor pressure of water
The Vapor Pressure of water depends on temperature, so the vapor pressure of water at room temperature (25°C) is 23.8mmHg. Factors affecting vapor pressure,
- Temperature: It has a high effect on vapor pressure. If thermal energy increases then vapor pressure increases, if thermal energy decreases vapor pressure decreases because thermal energy gives kinetic energy to the molecules of matter.
- Intermolecular Forces: The forces between molecules affect the vapor pressure.
Factors that do not affect vapor pressure,
- Quantity of liquid: Vapor pressure doesn’t depend on the quantity of substance. For example, the vapor pressure of a drop of water is equal to the vapor pressure of a jug of water.
- Surface area: Surface area affects the rate of evaporation it doesn’t affect vapor pressure.
How to calculate Vapor pressure using Raoult’s law
Raoult’s law relates the vapor pressure of solution (Psolution) to the vapor pressure of solvent (Psolvent) and mole fraction of solvent (Xsolvent).
Psolution = Psolvent × Xsolvent
Sample Problems
Question 1: Find the vapor pressure of a solution of syrup consisting of 1-liter water and 1-liter sucrose?
Solution:
First, find the mole fraction of the solvent. Here, the solvent is water.
Mass of 2 liter of water = 1000 × 2 = 2000gram
Mass of 1 liter of sucrose = 1056.7 × 2 = 2113.4gram
Mole of water = 2000 grams × 1 mol/18.015gram = 111.01859mole
Mole of sucrose = 2113.4gram × 1 mol/342.2965 g = 6.174179mole(using the molar mass of sucrose from chemical formula, C12H22O11)
Total mole = 111.01859 + 6.174179 = 117.2557694mole
Mole fraction of water = 111.01859/117.2557694 = 0.9468
Now, find the solvent vapor pressure? The vapor pressure of water at 25°C is 23.8 mm Hg.
Now, the values into Raoult’s law,
Psolution = Psolvent × Xsolvent
Psolution = (23.8 mm Hg)(0.9468)
Psolution = 22.54 mm Hg
Question 2: Measure the decrease in the vapor pressure when 0.003 kg of a solute that has a molecular mass of 57 is dissolved in 0.5 kg of water. The vapor pressure of water is 1.5 × 105 Pa.
Solution:
From question we have to find what is decrease in vapor pressure of solute that is dissolved in water,
Vapor pressure of solvent that is water = pº = 1.5 × 105 Pa,
Mass of water=W1= solvent= 0.5kg,
Mass of solute = W₂ = 0.003g,
Molecular mass of water = M₁ = 18 g mol-1,
Molecular mass of solute = M₂ = 57 g mol-1.
Decrease in vapor pressure =Po-P
(Po – P)/(Po) = (W2M1)/(W1M2)
(Po – P) = W2M1 × Po/(W1M2)
(Po – P) = ((0.003) × (18) × (1.5 × 105))/0.5 × 57
(Po – P) = 284.2105Pa
Hence, the decrease in vapor pressure is 284.2105Pa.
Question 3: What is the vapor pressure of a solution that has a solute mole fraction of 0.3? The vapor pressure of water is 16.358 mmHg at 23 °C.
Solution:
As per question here, we need to find vapor pressure of solute that is present in water
Xsolvent = 1.0000 − 0.3 = 0.7
From Raoult’s Law,
Psolution = (Xsolvent) (Psolvent)
x = (0.7) (16.358)
x = 11.4506 mmHg
Therefore, vapor pressure of solute is 11.4506mmHg
Question 4: The vapor pressure of water is 26 mm Hg at 30°C.Get the vapor pressure of the solution having 5 grams of urea (NH₂CONH₂) in 100 ml of water.
Solution:
Here, we need to find the vapor pressure of urea corresponding to the water vapor pressure.
Temperature of water = 30°C
Vapor pressure of solvent (water) = Po = 26 mm of Hg,
Mass of water (solvent) = W1 = 100ml=100gram
Mass of urea (solute)(NH₂CONH₂) = W₂ = 5gram,
Vapour pressure of solution = p =?
The molecular mass of urea (NH2CONH2) = M2
M₂ = 14 gram x 2+1gramx4+12gramx1+16 gram x 1 = 60 gram mol-1.
The molecular mass of water M₁ = 18 gram mol-1
(Po-P)/(Po) = (W2M1)/(W1M2)
(Po – P) = (W2M1) × Po/(W1M2)
P = (Po) – (W2M1) /(W1M2) × Po
P = (1- (5gram × 18 gram mol-1)/(100 × 60 gram mol-1)) × Po
P = (1- (5gram × 18 gram mol-1)/(100 × 60 gram mol-1)) × 26
P = (1 – (0.015)) × 26
P = 0.985 × 26
P = 25.61 mm of Hg
Therefore vapor pressure of solution is 25.61 mm of Hg
Question 5: The vapor pressure of water is 23.8 mmHg at 25°C. What is the vapor pressure of 3 molarity of C6H12O6?
Solution:
Here, we need to find vapor pressure of C6H12O6
Step 1: Conversion of molality to mole fraction.
Calculate total moles,
3 molarity of C6H12O6 = 3 mole / 1 kg H2O
1000 g / 18.015 g/mol = 55.51 mol
55.51 mole + 3 mole = 58.51 mole
Step 2: We need the mole fraction of water,
55.51 / 58.51 = 0.948
Step 3: Use Raoult’s law,
Psolution = (Xsolvent) (Psolvent)
Psolution = (0.948) (23.8 mmHg) = 22.579mmHg
Question 6: Solution of 400 g of sucrose in 600g of water at 30 °C. What is the vapor pressure of this solution? The vapor pressure of water is 31.82 mmHg at 30.0 °C
Solution:
Step 1: Getting number of mole of solute
400g / 342.2948 g/mol = 1.16858 mol
600 g / 18.015 g/mol = 33.305578 mol
Step 2: Getting mole fraction of the solvent,
33.305578 mol / (33.305578 mol + 1.16858 mol) = 0.9661
Step 3: Getting vapor pressure,
x = (33.305578 mmHg) (0.9661) = 32.176609 mmHg
Hence the vapor pressure solution is 32.176609 mmHg
Question 7: The vapor pressure of a solution is 23 mmHg at 28 °C. What is the mole fraction of solute in this solution? The vapor pressure of water is 26 mm Hg at 28 °C.
Solution:
From given, we have to find mole fraction of solute,
Use Raoult’s Law:
Psolution = (Xsolvent) (Psolvent)
23 = (x) (26)
Xsolvent = 0.88461538
0.88461538 is solvent mole fraction
Xsolute = 1 − 0.88461538 = 0.11538
Xsolute = 0.11538
0.11538 is solute mole fraction
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