Physics / Saturated Vapor Pressure
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EXPERIMENT 34 · PHYSICS LABORATORY

Saturated Vapor Pressure theory

Understand saturated vapor pressure: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
\ln(P) = -\frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T}\right) + C \iff y = m x + b \iff \Delta H_{\text{vap}} = -m R

Keep the physical quantities, graph variables and units consistent.

01Molecular Kinetics of Vaporization and Saturated Vapor Pressure

At the microscopic level, molecules in a liquid are bound together by intermolecular attractive forces (van der Waals dispersion, dipole-dipole interactions, and hydrogen bonding). Because of continuous thermal collisions, molecular kinetic energies are distributed according to the Maxwell-Boltzmann distribution: f(E) ∝ √E exp(−E / k_B T). Only those molecules located near the liquid surface whose normal kinetic energy component exceeds the cohesive binding energy barrier (the activation energy of vaporization E_vap) can overcome intermolecular attractions and escape into the vapor phase. The rate of vaporization per unit surface area is therefore proportional to exp(−E_vap / k_B T). Conversely, molecules in the vapor phase undergo chaotic thermal motion and frequently strike the liquid surface, where a fraction α_c (the condensation accommodation coefficient) is captured back into the liquid. The rate of condensation is directly proportional to the collision frequency of vapor molecules, which by the kinetic theory of gases is given by the Hertz-Knudsen relation: J_cond = P / √(2π m k_B T). When dynamic equilibrium is established, the rate of vaporization equals the rate of condensation: J_vap = J_cond. Equating these rates yields the equilibrium vapor pressure: P_sat ∝ T^(1/2) exp(−E_vap / k_B T), showing that vapor pressure increases exponentially with temperature. Because the equilibrium condition depends solely on the balance between molecular kinetic energy and intermolecular cohesive potential, the saturated vapor pressure is strictly an intensive thermodynamic property; it depends exclusively on temperature and the chemical identity of the liquid, remaining completely independent of the surface area or the relative volume of the container.

J_{\text{vap}} = J_{\text{cond}} = \frac{P_{\text{sat}}}{\sqrt{2\pi m k_B T}} \implies P_{\text{sat}}(T) = P_0 \exp\left(-\frac{\Delta H_{\text{vap}}}{R T}\right)
Why does the saturated vapor pressure of a liquid depend exponentially on temperature rather than linearly?

Vaporization requires molecules to overcome a cohesive energy barrier (E_vap). According to the Maxwell-Boltzmann distribution, the fraction of molecules possessing kinetic energy greater than this barrier scales exponentially as exp(−E_vap / k_BT). Raising the temperature increases this high-energy population exponentially, causing vapor pressure to rise exponentially.

02The Clapeyron and Clausius-Clapeyron Equations of State

The rigorous thermodynamic description of first-order phase transitions originates from the requirement of chemical potential equality across phase boundaries. For liquid and vapor phases in coexistence at temperature T and pressure P, the molar Gibbs free energies must be identical: G_m,l(T, P) = G_m,v(T, P). Moving along the coexistence curve by an infinitesimal increment (dT, dP) maintains equilibrium: dG_m,l = dG_m,v. Using the fundamental thermodynamic relation dG_m = −S_m dT + V_m dP, we obtain −S_m,l dT + V_m,l dP = −S_m,v dT + V_m,v dP, which rearranges to dP/dT = (S_m,v − S_m,l) / (V_m,v − V_m,l). Because the phase transition occurs reversibly at constant temperature, the molar entropy change is related to the molar latent heat of vaporization by ΔS_m = S_m,v − S_m,l = ΔH_vap / T. Substituting this into the derivative yields the exact Clapeyron equation: dP/dT = ΔH_vap / [T (V_m,v − V_m,l)]. Rudolf Clausius introduced two brilliant and physically justified approximations: (1) at pressures well below the critical point, the molar volume of the liquid is negligible compared to that of the vapor (V_m,v ≫ V_m,l, hence V_m,v − V_m,l ≈ V_m,v); (2) the vapor behaves as an ideal gas obeying V_m,v = RT/P. Substituting these approximations into Clapeyron’s equation gives dP/dT = (ΔH_vap P) / (R T²), which can be written in logarithmic form as d(ln P) / dT = ΔH_vap / (R T²), or equivalently, d(ln P) / d(1/T) = −ΔH_vap / R. Assuming ΔH_vap is constant over the measurement range, integration yields ln(P) = −(ΔH_vap / R)(1/T) + C, providing a powerful linear relationship that allows the enthalpy of vaporization to be extracted directly from experimental P-T data.

\frac{dP}{dT} = \frac{\Delta H_{\text{vap}}}{T (V_{m,v} - V_{m,l})} \approx \frac{\Delta H_{\text{vap}} P}{R T^2} \implies \frac{d(\ln P)}{d(1/T)} = -\frac{\Delta H_{\text{vap}}}{R}
What two key physical approximations transform the exact Clapeyron equation into the Clausius-Clapeyron equation?

First, neglecting the molar volume of the liquid compared to that of the vapor (V_m,v − V_m,l ≈ V_m,v). Second, treating the vapor as an ideal gas obeying V_m,v = RT/P. Both approximations are highly accurate for volatile liquids at temperatures well below their critical points.

03Intermolecular Forces and Hydrogen Bonding in Ethyl Alcohol

The magnitude of the enthalpy of vaporization ΔH_vap provides direct empirical quantification of the strength of intermolecular forces holding the liquid phase together. Ethyl alcohol (ethanol, CH₃CH₂OH) is an amphiphilic molecule containing both an ethyl non-polar hydrocarbon group and a polar hydroxyl (−OH) functional group. The electronegativity difference between oxygen (3.44) and hydrogen (2.20) creates a strong electric dipole moment along the O−H bond, enabling ethanol molecules to engage in extensive intermolecular hydrogen bonding. In liquid ethanol at room temperature, molecules form dynamic, transient oligomeric chains and rings held together by hydrogen bonds with individual bond dissociation energies of approximately 20 to 25 kJ mol⁻¹. In addition to hydrogen bonding, ethanol molecules experience dipole-dipole interactions and London dispersion forces arising from the polarizability of the electron cloud of the ethyl group. The total molar enthalpy of vaporization ΔH_vap ≈ 42 kJ mol⁻¹ represents the total energetic cost required to completely disrupt all these intermolecular attractive interactions and transfer one mole of ethanol from the condensed liquid state into the non-interacting vapor state. In contrast, non-polar liquids of comparable molecular weight (such as propane, CH₃CH₂CH₃, M = 44 g mol⁻¹, ΔH_vap ≈ 19 kJ mol⁻¹) exhibit much weaker dispersion forces and boil at −42 °C. Diethyl ether ((C₂H₅)₂O, M = 74 g mol⁻¹), which has dipole-dipole forces but lacks hydrogen-bond donors, has ΔH_vap ≈ 27 kJ mol⁻¹ and boils at 34.6 °C. The substantially higher enthalpy of vaporization and elevated boiling point of ethanol (78.4 °C) provide striking thermodynamic evidence for the robust network of intermolecular hydrogen bonds in the liquid.

\Delta H_{\text{vap}} = \Delta H_{\text{dispersion}} + \Delta H_{\text{dipole}} + \Delta H_{\text{H-bond}} \approx 42\text{ kJ mol}^{-1} \text{ (for C}_2\text{H}_5\text{OH)}
Why is the enthalpy of vaporization of ethanol (42 kJ/mol) substantially higher than that of diethyl ether (27 kJ/mol) despite ether having a larger molecular mass?

Ethanol possesses a hydroxyl (−OH) group capable of forming strong intermolecular hydrogen bonds, whereas diethyl ether has no O−H bonds and interacts only through weaker dipole-dipole and dispersion forces. Breaking the hydrogen-bonding network in ethanol requires significantly greater thermal energy.

04Boiling Point Thermodynamics and Ambient Barometric Dependence

The boiling point of a liquid is thermodynamically defined as the temperature at which its saturated vapor pressure equals the prevailing external atmospheric pressure: P_sat(T_b) = P_ext. At temperatures below the boiling point, vaporization is strictly confined to the liquid surface because any vapor bubble attempting to form within the bulk liquid would have an internal pressure P_sat < P_ext and would be instantaneously crushed by the surrounding hydrostatic pressure. When the liquid is heated to T_b such that P_sat = P_ext, vapor bubbles can nucleate and grow throughout the bulk volume, leading to the rapid, vigorous phase transition known as boiling. The normal boiling point is specifically defined for an external pressure of exactly 1 standard atmosphere (P_ext = 101.325 kPa = 760.0 mmHg), which for pure ethanol is 78.37 °C. Because ambient atmospheric pressure fluctuates with weather systems and decreases exponentially with geographic altitude, the observed boiling point varies from day to day. The sensitivity of the boiling point to barometric pressure is quantitatively predicted by the inverted Clausius-Clapeyron equation: dT_b / dP = (R T_b²) / (ΔH_vap P). For ethanol at 1 atm, dT_b / dP ≈ 0.034 °C per mmHg (or 0.255 °C per kPa). A drop in barometric pressure of 20 mmHg (common during cyclonic weather) depresses the boiling point of ethanol by nearly 0.7 °C. Trouton’s rule provides an empirical benchmark for non-associated liquids, stating that the entropy of vaporization at the normal boiling point is approximately constant: ΔS_vap = ΔH_vap / T_b ≈ 85 to 88 J mol⁻¹ K⁻¹. For ethanol, ΔS_vap = 41,920 J mol⁻¹ / 351.52 K ≈ 119.2 J mol⁻¹ K⁻¹, which significantly exceeds Trouton’s constant. This positive deviation occurs because liquid ethanol is highly ordered due to hydrogen bonding; vaporizing it destroys this microscopic order, resulting in a substantially larger increase in entropy.

\frac{dT_b}{dP} = \frac{R T_b^2}{\Delta H_{\text{vap}} P}, \quad \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_b} \approx 119.2\text{ J mol}^{-1}\text{ K}^{-1} > 88\text{ J mol}^{-1}\text{ K}^{-1} \text{ (Trouton deviation)}
Why does ethanol exhibit a significant positive deviation from Trouton’s rule (ΔS_vap ≈ 119 J/(mol·K) vs 88 J/(mol·K))?

Trouton’s rule applies to disordered liquids. Liquid ethanol has a high degree of microscopic structural order resulting from its hydrogen-bonded network. Vaporization disrupts this ordered structure, causing an exceptionally large gain in entropy upon entering the random gas phase.

05Empirical Vapor Pressure Equations: Antoine and Wagner Formulations

Over wide temperature intervals extending from the triple point to the critical point, the assumption of constant enthalpy of vaporization breaks down because the heat capacity of the vapor differs from that of the liquid: d(ΔH_vap)/dT = ΔC_p = C_p,v − C_p,l. Because the liquid has additional vibrational and librational modes, C_p,l > C_p,v, meaning ΔC_p is negative and ΔH_vap decreases steadily as temperature rises toward the critical temperature T_c (where ΔH_vap vanishes identically). To achieve high-precision representation of experimental vapor pressures across engineering applications, empirical equations with additional fitting parameters are employed. The most widely used formulation is the three-parameter Antoine equation: log₁₀(P) = A − B / (T + C), where A, B, and C are substance-specific empirical constants and T is temperature in Celsius. For ethyl alcohol over the temperature range of 20 °C to 90 °C, the standard Antoine parameters (with P in mmHg and T in °C) are A = 8.20417, B = 1642.89, and C = 230.300. The parameter C acts as a temperature adjustment that accounts for the curvature induced by ΔC_p. In thermodynamic metrology, the even more sophisticated four-parameter Wagner equation is used for reference-grade data: ln(P / P_c) = (T_c / T) [a τ + b τ^1.5 + c τ^3 + d τ^6], where τ = 1 − T/T_c. However, over the moderate 60 °C laboratory interval investigated here (20 °C to 80 °C), the simple two-parameter Clausius-Clapeyron equation fits the experimental data with an extraordinary coefficient of determination (R² > 0.9999), demonstrating that the mean effective enthalpy of vaporization ΔH_vap is an exceptionally accurate thermodynamic parameter for instructional physics.

\log_{10}(P_{\text{mmHg}}) = A - \frac{B}{T + C} \iff \ln(P_{\text{Pa}}) \approx -\frac{\Delta H_{\text{eff}}}{R}\left(\frac{1}{T}\right) + C_0
Why does the enthalpy of vaporization ΔH_vap decrease at higher temperatures and vanish at the critical point?

As temperature rises, thermal expansion diminishes the density of the liquid while compression increases the density of the vapor. At the critical point, the liquid and vapor densities become identical and the meniscus disappears, meaning no energetic barrier exists between phases, so ΔH_vap becomes zero.

06Manometric Metrology: Saturated Vapor Tubes and Volatile Mixtures

The physical measurement of saturated vapor pressure requires isolating the liquid-vapor system from non-condensable ambient gases. In the classical laboratory method for saturated vapors, pairs of vapor pressure tubes are utilized: one closed (evacuated and sealed) and one open to atmospheric pressure. The closed tube operates as a differential barometer: the space above the mercury column contains only the volatile liquid and its saturated vapor. When the tube is submerged in a cold bath, the vapor pressure is low and the mercury column stands high; as the bath is warmed, the vapor pushes the mercury column down. In an open-tube manometer, the volatile liquid is contained inside a bulb connected to a U-tube containing mercury whose outer limb is open to the atmosphere. The absolute pressure is then P = B + (h_open − h_closed). A crucial precaution is degassing the liquid before sealing: any dissolved atmospheric air (oxygen or nitrogen) will desorb into the vapor space upon heating. In accordance with Dalton’s law of partial pressures, the total measured pressure would be P_total = P_sat + P_air, producing a false positive pressure offset that corrupts the Clausius-Clapeyron slope. In binary mixtures (such as an ethanol-ether system), Raoult’s law states that each volatile component exerts a partial pressure proportional to its liquid mole fraction: P_i = x_i P_i*. Because diethyl ether has a much higher vapor pressure than ethanol, the vapor phase of the mixture is strongly enriched in ether, illustrating the fundamental principle of fractional distillation.

P_{\text{total}} = P_{\text{sat}}(T) + P_{\text{air}}, \quad P_{\text{mixture}} = x_A P_A^* + x_B P_B^* \text{ (Raoult’s Law)}
How does dissolved air in the volatile liquid introduce a systematic error into the measured vapor pressure?

Dissolved air desorbs from the liquid upon heating and enters the vapor headspace. By Dalton’s law of partial pressures, the manometer measures P_total = P_sat + P_air. Because the trapped air pressure increases with temperature according to the ideal gas law, it adds a temperature-dependent offset that distorts both the intercept and the slope of the Clausius-Clapeyron fit.

07Error Propagation and Linear Regression in Clausius-Clapeyron Analysis

To extract the enthalpy of vaporization ΔH_vap and its experimental uncertainty from paired measurements of temperature t and vapor pressure P, the data are transformed into linear coordinates: x_i = 1 / T_i (where T_i = t_i + 273.15 K) and y_i = ln(P_i / Pa). The ordinary least-squares regression line y = m x + b is fitted to the n data points. The slope m is given by m = [n Σ(x_i y_i) − (Σx_i)(Σy_i)] / [n Σx_i² − (Σx_i)²]. The molar enthalpy of vaporization is directly proportional to the slope: ΔH_vap = −m R. Because R is an exact fundamental physical constant defined in the SI system (R = 8.314462618 J mol⁻¹ K⁻¹), the standard error in the enthalpy of vaporization is determined entirely by the standard error of the regression slope: s_{ΔH} = R · s_m = R · √[s² / Σ(x_i − x̄)²], where s² = Σ(y_i − mx_i − b)² / (n − 2) is the residual variance of the fit. A critical metrological consideration is that the logarithmic transformation y = ln(P) non-linearly weights experimental pressure errors: dy = dP / P. A constant absolute pressure measurement uncertainty (e.g., ±0.2 mmHg from scale reading) translates into a relative uncertainty dy = δP / P that is substantially larger at low temperatures where P is small (at 20 °C, P ≈ 44 mmHg, so dy ≈ 0.2/44 = 0.0045) than at high temperatures near the boiling point (at 80 °C, P ≈ 812 mmHg, so dy ≈ 0.2/812 = 0.00025). High-precision analytical evaluations may utilize weighted least-squares regression with weights w_i = 1 / (δy_i)² ∝ P_i²; however, over the five standard stations investigated here, ordinary least squares yields an estimate within 0.02% of the reference enthalpy, demonstrating exceptional metrological robustness.

y = \ln(P), \quad x = \frac{1}{T}, \quad m = -\frac{\Delta H_{\text{vap}}}{R} \implies s_{\Delta H} = R \cdot s_m = R \sqrt{\frac{s^2}{\sum (x_i - \bar{x})^2}}
Why does a constant absolute pressure error δP have a much larger impact on the Clausius-Clapeyron variable y = ln(P) at low temperatures than at high temperatures?

Because dy = d(ln P) = dP / P. The error in y is inversely proportional to the absolute pressure P. At low temperatures, P is small, so a given pressure error δP constitutes a much larger fractional uncertainty, producing a larger deviation in y.

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Connect the equations to a measurement.

The Clausius-Clapeyron equation provides the thermodynamic cornerstone connecting macroscopic phase equilibria with microscopic intermolecular binding forces. The measured molar enthalpy of vaporization ΔH_vap = 41.92 kJ mol⁻¹ demonstrates the substantial energetic cost of disrupting intermolecular hydrogen bonds during the liquid-to-vapor phase transition.