Best real-world examples of calculating vapor pressure of solutions

Why start with examples of calculating vapor pressure of solutions?

Raoult’s law is simple on paper:

Pᵢ = xᵢ · Pᵢ°

where Pᵢ is the partial vapor pressure of component i, xᵢ is its mole fraction in the liquid, and Pᵢ° is its pure-component vapor pressure at that temperature.

But where students actually learn it is in examples of calculating vapor pressure of solutions with real numbers, real units, and real mixtures that don’t always behave perfectly. So we’ll build from nearly ideal systems to more realistic ones, including electrolytes and non‑ideal organic mixtures.

Classic organic mixture: benzene–toluene as an example of calculating vapor pressure

A textbook favorite: a liquid solution of benzene and toluene at 25 °C. These two are chemically similar and behave close to ideal, making them a clean example of calculating vapor pressure of solutions.

Data (approximate, 25 °C):

• Pure benzene vapor pressure, P°(benzene) ≈ 100 mmHg
• Pure toluene vapor pressure, P°(toluene) ≈ 30 mmHg

Say the liquid mixture is 0.40 mole fraction benzene and 0.60 mole fraction toluene.

Partial pressures:

• P(benzene) = x(benzene) · P°(benzene)
  = 0.40 × 100 mmHg
  = 40 mmHg

• P(toluene) = x(toluene) · P°(toluene)
  = 0.60 × 30 mmHg
  = 18 mmHg

Total vapor pressure:

P_total = P(benzene) + P(toluene) = 40 + 18 = 58 mmHg

This is one of the best examples for beginners because:

• It’s nearly ideal, so Raoult’s law works very well.
• The more volatile component (benzene) dominates the vapor phase.
• It directly connects to distillation and separation processes taught in chemical engineering.

Aqueous sugar solution: low volatility solute example of calculating vapor pressure

Now consider a beaker of water with dissolved sucrose (table sugar). Sugar is effectively nonvolatile at room temperature, so only water contributes to the vapor phase.

Suppose you dissolve 0.50 mol of sucrose in 1.50 mol of water at 25 °C.

• Total moles in the liquid = 0.50 + 1.50 = 2.00 mol
• Mole fraction of water: x(water) = 1.50 / 2.00 = 0.75

At 25 °C, the vapor pressure of pure water is about 23.8 mmHg.

Using Raoult’s law:

P(water) = x(water) · P°(water)
= 0.75 × 23.8 mmHg
≈ 17.9 mmHg

Total vapor pressure above the solution is about 17.9 mmHg, lower than pure water.

This example of calculating vapor pressure of solutions shows why:

• Adding nonvolatile solute lowers vapor pressure.
• Foods with high sugar content dry out more slowly.
• Vapor pressure lowering connects directly to boiling point elevation and freezing point depression (classic colligative property territory, well covered in general chemistry texts such as those from MIT OpenCourseWare and other .edu resources).

Saltwater and electrolytes: real examples of calculating vapor pressure of solutions

Salt doesn’t just dissolve; it dissociates. A 1 mol solution of NaCl in water behaves closer to 2 mol of particles (Na⁺ and Cl⁻). That influences vapor pressure lowering.

Imagine a solution at 25 °C containing 1.0 mol NaCl dissolved in 9.0 mol water.

If we treat NaCl as fully dissociated into 2 ions per formula unit, we have:

• Water: 9.0 mol
• Ions: 1.0 mol NaCl × 2 ≈ 2.0 “solute particles”

Total effective particles ≈ 9.0 + 2.0 = 11.0

Approximate mole fraction of water:

x(water) ≈ 9.0 / 11.0 ≈ 0.82

Again using P°(water) ≈ 23.8 mmHg at 25 °C:

P(solution) ≈ 0.82 × 23.8 mmHg ≈ 19.5 mmHg

This is an oversimplified but instructive example of calculating vapor pressure of solutions containing electrolytes. It explains why:

• Seawater evaporates more slowly than pure water.
• Salt is used in food preservation; lower water activity limits microbial growth (see discussions of water activity in food safety on sites like the USDA or FDA).

For more rigorous treatment, chemical engineers use activity coefficients rather than assuming ideal behavior. The U.S. National Institute of Standards and Technology (NIST) provides thermodynamic data and models that go beyond basic Raoult’s law for these systems:
https://www.nist.gov

Antifreeze in car radiators: glycol–water as one of the best examples

Ethylene glycol or propylene glycol in water is a very practical example of calculating vapor pressure of solutions. Lower vapor pressure means a higher boiling point, which is exactly what you want in an engine cooling system.

Take a 50/50 by mole ethylene glycol–water mixture at 100 °C. At 100 °C, the vapor pressure of pure water is 760 mmHg (1 atm). Ethylene glycol has a much lower vapor pressure at this temperature, and for a first-pass classroom example, it’s often treated as nonvolatile.

Suppose:

• x(water) = 0.50
• x(glycol) = 0.50
• P°(water, 100 °C) = 760 mmHg
• P°(glycol, 100 °C) ≈ 0 (assumed for simplicity)

Then:

P(water) = 0.50 × 760 mmHg = 380 mmHg

P_total ≈ 380 mmHg

So at 100 °C, the solution’s vapor pressure is only about half an atmosphere. That means the solution will not boil at 100 °C; it needs a higher temperature to reach 1 atm vapor pressure. This is why a 50/50 antifreeze mix raises the boiling point of coolant, as discussed in automotive maintenance resources and engineering courses.

As engine technology evolves, manufacturers keep optimizing coolant formulations for safety and environmental impact, but the physics under the hood still rests on the same Raoult’s law logic.

Non-ideal behavior: ethanol–water as a real example of vapor pressure calculations

Ethanol–water is the poster child for non‑ideal mixtures. It forms an azeotrope, which means the vapor and liquid compositions match at a particular mixture and temperature. Still, you can treat it as ideal over narrow composition ranges to get a feel for the numbers.

Take an ethanol–water solution at 40 °C. Approximate pure-component vapor pressures:

• P°(ethanol, 40 °C) ≈ 135 mmHg
• P°(water, 40 °C) ≈ 55 mmHg

Say the liquid is 0.30 mole fraction ethanol and 0.70 mole fraction water.

Partial pressures:

• P(ethanol) = 0.30 × 135 ≈ 40.5 mmHg
• P(water) = 0.70 × 55 ≈ 38.5 mmHg

Total pressure:

P_total ≈ 40.5 + 38.5 = 79 mmHg

Even with this simple Raoult’s law approach, you see ethanol’s higher volatility. In reality, hydrogen bonding and non‑ideal interactions shift these values. For deeper treatment, thermodynamics courses and research articles from universities like MIT or UC Berkeley (both .edu) discuss activity coefficient models (e.g., Wilson, NRTL, UNIQUAC) that correct Raoult’s law for non‑ideal systems.

Industrial air–water systems: humidity as an example of calculating vapor pressure of solutions

Humidity problems are, quietly, examples of calculating vapor pressure of solutions too. Air with water vapor can be described with partial pressures and mole fractions.

At 25 °C, pure water has a vapor pressure of about 23.8 mmHg. If the air is at 50% relative humidity, the partial pressure of water vapor in the air is:

P(H₂O, air) = 0.50 × 23.8 mmHg ≈ 11.9 mmHg

Now imagine a cooling tower where warm process water contacts air. Engineers estimate how much water will evaporate by comparing the vapor pressure of water at the liquid temperature to the partial pressure in the air. The driving force is the difference between those two.

This is how you connect Raoult’s law–style thinking to:

• Building HVAC design
• Industrial dryers
• Cooling towers and evaporative coolers

Psychrometric charts used in mechanical engineering essentially encode examples of calculating vapor pressure of solutions (water in air) across a wide range of temperatures and humidities. ASHRAE and engineering departments at universities like Penn State or Georgia Tech provide educational material on this topic.

Pharmaceutical and food systems: real examples include solvent blends and stabilizers

Modern pharma and food science use Raoult’s law ideas constantly, especially when designing solvent systems or controlling moisture.

One classic pharmaceutical example of calculating vapor pressure of solutions is a binary solvent blend used to control evaporation rate in a coating process. Imagine a tablet coating solution with 0.60 mole fraction isopropanol and 0.40 mole fraction water at 30 °C.

Approximate pure-component vapor pressures at 30 °C:

• P°(isopropanol) ≈ 45 mmHg
• P°(water) ≈ 31.8 mmHg

Partial pressures:

• P(IPA) = 0.60 × 45 ≈ 27 mmHg
• P(water) = 0.40 × 31.8 ≈ 12.7 mmHg

Total:

P_total ≈ 39.7 mmHg

Engineers adjust the mole fractions to tune how fast the solvent mixture evaporates from the tablet surface. Too fast and you get cracked coatings; too slow and production bottlenecks.

In food science, vapor pressure lowering is tied directly to water activity, a key metric for microbial growth risk. The U.S. Department of Agriculture (USDA) and FDA both publish guidance on how water activity and solute content (sugars, salts) impact food safety:

• USDA Food Safety and Inspection Service: https://www.fsis.usda.gov
• U.S. Food and Drug Administration: https://www.fda.gov

These are not just abstract ideas; they’re real examples where the same Raoult’s law math underpins shelf life and safety decisions.

Beyond ideal Raoult’s law: 2024–2025 trends in modeling vapor pressure

In 2024–2025, the theory hasn’t changed, but the tools have. Chemists and engineers still teach Raoult’s law as the starting point, but now it’s routinely wrapped inside software and databases:

• Process simulators (Aspen Plus, CHEMCAD, HYSYS) embed Raoult’s law with activity coefficient models for real mixtures.
• NIST’s Chemistry WebBook and related databases provide updated vapor pressure and thermodynamic data: https://webbook.nist.gov
• Open‑source tools like DWSIM and Python libraries (e.g., CoolProp) let students and researchers run vapor–liquid equilibrium calculations with modern equations of state.

The big picture: Raoult’s law remains the first step in understanding, and examples of calculating vapor pressure of solutions are still the main way students internalize it. But in practice, professionals quickly layer on:

• Activity coefficients (γᵢ) to correct for non‑ideal behavior
• Temperature dependence of P°(T) using Antoine or other correlations
• Multicomponent mixtures far more complex than the two‑component examples you see in class

If you’re heading into research or industry, getting comfortable with hand calculations on simple examples makes it much easier to trust—and question—what the software spits out.

FAQ: Short answers built around real examples

Q1: Can you give a simple example of calculating vapor pressure of solutions for an exam?
Yes. A classic exam problem is a benzene–toluene mixture at a fixed temperature. You’re given pure-component vapor pressures and liquid mole fractions, then asked for partial and total pressures. That benzene–toluene case we walked through earlier is almost exactly what shows up on many general chemistry and physical chemistry tests.

Q2: How do I handle examples of solutions with nonvolatile solutes like salt or sugar?
Treat the solute as having zero vapor pressure and focus on the solvent. Compute the solvent mole fraction in the liquid, then multiply by the pure solvent vapor pressure. That’s why the sugar–water and NaCl–water examples above are so common: they show how adding solute lowers the solvent’s vapor pressure and connects directly to colligative properties.

Q3: Are there good real‑life examples of calculating vapor pressure of solutions outside the lab?
Absolutely. Antifreeze mixtures in car radiators, solvent blends in paints and coatings, humidity control in buildings, and food preservation through salt or sugar addition are all practical examples. In each case, engineers are effectively managing vapor pressure by adjusting composition.

Q4: How accurate is Raoult’s law for real mixtures?
It works best for ideal or nearly ideal solutions: similar molecules, no strong specific interactions, moderate concentrations. For systems like ethanol–water, strongly hydrogen‑bonded or highly non‑ideal solutions, you need activity coefficients and more advanced models. University thermodynamics courses and NIST resources go into the details with data‑backed models.

Q5: Where can I find more data to practice examples of vapor pressure calculations?
For pure-component vapor pressures and thermodynamic properties, the NIST Chemistry WebBook is a widely used reference: https://webbook.nist.gov. Many university chemistry departments (.edu) also host tables and example problem sets that mirror the kinds of calculations shown here.