Best examples of vapor pressure calculation examples for mixtures (step‑by‑step)

Why start with examples of vapor pressure calculation examples for mixtures?

Textbook definitions of Raoult’s law are fine, but the real learning happens when you push through multiple, varied examples. When you compare several examples of vapor pressure calculation examples for mixtures side by side, patterns jump out:

• How mole fraction directly scales partial vapor pressures.
• Why total vapor pressure is just the sum of component contributions.
• Where ideal behavior breaks down and activity coefficients sneak in.

Below, we’ll build up a small “toolkit” of examples, from simple binary ideal mixtures to more realistic cases that connect to modern chemical engineering and environmental topics.

Example of an ideal binary mixture: benzene–toluene at 25 °C

This is the classic classroom case because benzene and toluene form an almost ideal solution. It’s one of the best examples of vapor pressure calculation examples for mixtures when you’re first learning Raoult’s law.

Given (at 25 °C, approximate values):

• Pure benzene vapor pressure, \(P^*_{\text{benzene}} \approx 95\,\text{mmHg}\)
• Pure toluene vapor pressure, \(P^*_{\text{toluene}} \approx 28\,\text{mmHg}\)
• Liquid mixture: 40 mol% benzene, 60 mol% toluene

So the liquid mole fractions are:

• \(x_{\text{benzene}} = 0.40\)
• \(x_{\text{toluene}} = 0.60\)

Raoult’s law for an ideal solution:

[
P_i = x_i P^*_i
]

Step 1: Partial vapor pressures

• Benzene: \(P_{\text{benzene}} = 0.40 \times 95 = 38\,\text{mmHg}\)
• Toluene: \(P_{\text{toluene}} = 0.60 \times 28 = 16.8\,\text{mmHg}\)

Step 2: Total vapor pressure

[
P_\text{total} = P_{\text{benzene}} + P_{\text{toluene}} = 38 + 16.8 = 54.8\,\text{mmHg}
]

Step 3: Vapor composition (using Dalton’s law):

[
y_i = \frac{P_i}{P_\text{total}}
]

• \(y_{\text{benzene}} = 38 / 54.8 \approx 0.69\)
• \(y_{\text{toluene}} = 16.8 / 54.8 \approx 0.31\)

So even though the liquid is only 40% benzene, the vapor is about 69% benzene because benzene is more volatile. This is one of the clearest examples of vapor pressure calculation examples for mixtures showing how distillation enriches the more volatile component.

Ethanol–water: a realistic, slightly non‑ideal mixture

If benzene–toluene is the tidy textbook case, ethanol–water is the messy real‑world one. It’s widely used in labs, industry, and beverages, and it exhibits non‑ideal behavior and an azeotrope.

Given (at 25 °C, approximate values):

• Pure water vapor pressure, \(P^*_{\text{water}} \approx 24\,\text{mmHg}\)
• Pure ethanol vapor pressure, \(P^*_{\text{ethanol}} \approx 59\,\text{mmHg}\)
• Liquid mixture: 30 mol% ethanol, 70 mol% water

If we pretend it’s ideal for a first‑pass calculation:

• \(x_{\text{ethanol}} = 0.30\)
• \(x_{\text{water}} = 0.70\)

Step 1: Ideal Raoult’s law prediction

• \(P_{\text{ethanol}} = 0.30 \times 59 \approx 17.7\,\text{mmHg}\)
• \(P_{\text{water}} = 0.70 \times 24 \approx 16.8\,\text{mmHg}\)

Total:

[
P_\text{total, ideal} \approx 34.5\,\text{mmHg}
]

In practice, measured vapor pressures deviate from this because of hydrogen bonding and non‑ideal interactions. Modern physical chemistry and chemical engineering courses often introduce activity coefficients here, using models like Wilson or NRTL, especially in 2024–2025 curricula that emphasize computational thermodynamics. For students, this mixture is one of the best examples of vapor pressure calculation examples for mixtures that bridge simple Raoult’s law and more advanced models.

For actual data tables and phase diagrams, you can check resources like NIST’s Chemistry WebBook: https://webbook.nist.gov/chemistry/

Example of a low‑volatility solute: salt water (NaCl in water)

Salt water is a nice contrast because the solute (NaCl) has essentially no vapor pressure. This is a classic example of Raoult’s law applied to a non‑volatile solute.

Given (25 °C):

• Pure water vapor pressure: \(P^*_{\text{water}} \approx 24\,\text{mmHg}\)
• Aqueous NaCl solution: 1.0 mol NaCl in 9.0 mol water

Assume complete dissociation is ignored for a first approximation (you can later refine with van ’t Hoff factors).

Total moles in liquid:

[
n_\text{total} = 1.0 + 9.0 = 10.0\,\text{mol}
]

Mole fraction of water:

[
x_{\text{water}} = \frac{9.0}{10.0} = 0.90
]

NaCl is non‑volatile, so it doesn’t contribute to vapor pressure. Apply Raoult’s law to water:

[
P_{\text{water}} = x_{\text{water}} P^*_{\text{water}} = 0.90 \times 24 \approx 21.6\,\text{mmHg}
]

The vapor pressure is lowered from 24 mmHg (pure water) to about 21.6 mmHg. This is more than just a homework exercise: vapor pressure lowering connects directly to boiling point elevation and freezing point depression, which show up in environmental topics and in resources like the EPA’s climate and atmospheric chemistry materials (https://www.epa.gov/).

Gasoline‑like mixture: hexane–heptane

Real fuels are complex mixtures, but a simple two‑component model like hexane–heptane still gives useful insight. In 2024–2025, engineering coursework often uses these kinds of examples to show how vapor pressure relates to volatility and emissions.

Given (25 °C, approximate):

• \(P^*_{\text{n‑hexane}} \approx 150\,\text{mmHg}\)
• \(P^*_{\text{n‑heptane}} \approx 45\,\text{mmHg}\)
• Liquid mixture: 50 mol% hexane, 50 mol% heptane

Mole fractions:

• \(x_{\text{hexane}} = 0.50\)
• \(x_{\text{heptane}} = 0.50\)

Step 1: Partial pressures

• \(P_{\text{hexane}} = 0.50 \times 150 = 75\,\text{mmHg}\)
• \(P_{\text{heptane}} = 0.50 \times 45 = 22.5\,\text{mmHg}\)

Total:

[
P_\text{total} = 75 + 22.5 = 97.5\,\text{mmHg}
]

Even in this very simple example of a hydrocarbon mixture, you can see that hexane dominates the vapor phase. When environmental agencies and health organizations describe volatile organic compounds (VOCs) and inhalation exposure, they’re implicitly talking about these kinds of vapor pressure behaviors. For general health‑oriented discussion of solvent exposure, sites like the Agency for Toxic Substances and Disease Registry (ATSDR, part of CDC): https://www.atsdr.cdc.gov/ are useful starting points.

Example of vapor pressure calculation examples for mixtures in distillation design

Now let’s connect the math to a process decision. Imagine you’re designing a simple batch distillation to separate benzene and toluene.

You start with the benzene–toluene mixture from earlier: 40 mol% benzene in the liquid at 25 °C, with total vapor pressure 54.8 mmHg and vapor composition ~69% benzene.

If you gently boil the liquid at a higher temperature (say, near 1 atm total pressure), the same Raoult’s law logic applies, just with new pure‑component vapor pressures at the elevated temperature. The more volatile benzene will still dominate the vapor phase. By repeatedly condensing that vapor, you get a benzene‑rich distillate.

In real design work, engineers pair many such examples of vapor pressure calculation examples for mixtures with equilibrium data and McCabe–Thiele diagrams. Even though industrial simulators now do the heavy lifting, the underlying Raoult’s law calculations are the first sanity check.

Non‑ideal behavior: activity coefficients in modern examples

By 2024–2025, most upper‑level courses and many industrial guidelines emphasize that Raoult’s law is an ideal model. Non‑ideal mixtures require a modified form:

[
P_i = x_i \gamma_i P^*_i
]

Here, \(\gamma_i\) is the activity coefficient. Mixtures like ethanol–water, acetone–chloroform, or many aqueous organics show \(\gamma_i \neq 1\).

As a quick, conceptual example of vapor pressure calculation examples for mixtures with non‑ideality, suppose you have an ethanol–water mixture where experimental data indicate an activity coefficient for ethanol of \(\gamma_{\text{ethanol}} = 1.3\) at a given composition and temperature.

Using the earlier ideal estimate for ethanol’s partial pressure (17.7 mmHg), you would correct it to:

[
P_{\text{ethanol, real}} = x_{\text{ethanol}} \gamma_{\text{ethanol}} P^*_{\text{ethanol}} = 1.3 \times 17.7 \approx 23.0\,\text{mmHg}
]

That higher partial pressure changes both the total vapor pressure and the vapor composition. Modern chemical engineering software (e.g., Aspen Plus, CHEMCAD) automates this, but students still work through hand‑calculated examples of vapor pressure calculation examples for mixtures to understand what the activity coefficient is actually doing.

For deeper background on phase equilibria and activity, university notes from sites like MIT OpenCourseWare (https://ocw.mit.edu/) or other .edu thermodynamics courses are good references.

Temperature dependence: using Antoine equations in worked examples

So far, we’ve treated pure‑component vapor pressures \(P^*_i\) as given. In real problems, you often have to **calculate** \(P^*_i\) from temperature using an Antoine equation:

[
\log_{10} P^* = A - \frac{B}{C + T}
]

with \(P^*\) in mmHg and \(T\) in °C.

Consider a new example of an ideal mixture: acetone–chloroform at 35 °C. Suppose Antoine constants (hypothetical here for illustration) give:

• \(P^*_{\text{acetone}}(35\,^{\circ}\text{C}) = 300\,\text{mmHg}\)
• \(P^*_{\text{chloroform}}(35\,^{\circ}\text{C}) = 200\,\text{mmHg}\)

Take a liquid mixture with 25 mol% acetone, 75 mol% chloroform.

Mole fractions:

• \(x_{\text{acetone}} = 0.25\)
• \(x_{\text{chloroform}} = 0.75\)

Step 1: Partial pressures

• \(P_{\text{acetone}} = 0.25 \times 300 = 75\,\text{mmHg}\)
• \(P_{\text{chloroform}} = 0.75 \times 200 = 150\,\text{mmHg}\)

Step 2: Total vapor pressure

[
P_\text{total} = 75 + 150 = 225\,\text{mmHg}
]

This illustrates how temperature and pure‑component data are woven into examples of vapor pressure calculation examples for mixtures. In real lab or process work, you would pull Antoine constants or direct vapor pressure data from trusted databases like NIST.

Environmental and health‑related examples include vapor pressure calculations

Vapor pressure is not just an exam topic; it shows up in environmental science and public health.

Paint solvents and indoor air quality

Consider a simplified paint thinner mixture of toluene and isopropanol. Higher vapor pressure components evaporate more readily and can affect indoor air quality. Estimating the partial pressures of each component helps predict airborne concentrations, which feeds into exposure assessments.

Health‑oriented organizations like the National Institute for Occupational Safety and Health (NIOSH, via CDC: https://www.cdc.gov/niosh/) and educational medical sites such as Mayo Clinic (https://www.mayoclinic.org/) often discuss solvent exposure qualitatively. Behind the scenes, the quantitative side uses the same style of Raoult’s law calculations you see in the best examples of vapor pressure calculation examples for mixtures.

Environmental fate of organic pollutants

Regulators and researchers use vapor pressure, Henry’s law constants, and related data to estimate how pollutants partition between water, soil, and air. For many organic contaminants, the starting point is still the pure‑component vapor pressure and its effective contribution in a mixture.

Pulling it together: patterns across all these examples

Looking back over these real examples, a few consistent themes stand out:

• In every ideal example of a binary liquid, the more volatile component is enriched in the vapor phase compared with the liquid.
• Non‑volatile solutes (like salts) lower the vapor pressure of the solvent in direct proportion to the solvent’s mole fraction.
• Non‑ideal behavior can be handled by introducing activity coefficients, but the basic structure of Raoult’s law remains.
• Temperature enters through pure‑component vapor pressures, often via Antoine equations.

If you’re studying for exams or working in a lab, the best way to lock this in is to work through many examples of vapor pressure calculation examples for mixtures: change the mole fractions, change the temperature, and see how the vapor composition responds. The math is repetitive, but the physical insight you gain is worth it.

FAQ: short answers based on common questions

What are some common examples of vapor pressure calculation examples for mixtures in exams?

Common textbook and exam problems often use benzene–toluene, hexane–heptane, and acetone–chloroform as ideal mixtures. Ethanol–water and salt solutions (NaCl in water) show up as classic non‑ideal and colligative property examples.

Can Raoult’s law be used as an example of predicting boiling points of mixtures?

Yes. Once you have the total vapor pressure as a function of temperature and composition, you can find the temperature where the total vapor pressure equals the external pressure (often 1 atm). That temperature is the boiling point of the mixture. Many distillation design problems are really extended examples of vapor pressure calculation examples for mixtures.

Is there an example of Raoult’s law failing badly?

Mixtures with strong specific interactions—like water with many organics, or systems with hydrogen bonding and association—can deviate strongly from Raoult’s law. Ethanol–water near the azeotropic composition is a well‑known example where ideal assumptions give poor predictions, and activity‑coefficient models are needed.

Where can I find more real examples and data tables for vapor pressure calculations?

For high‑quality data, NIST’s Chemistry WebBook (https://webbook.nist.gov/chemistry/) is widely used. For educational explanations, .edu thermodynamics and physical chemistry course pages (for example, MIT OpenCourseWare) provide many worked examples of vapor pressure calculation examples for mixtures.