Vapor Pressure of a Pure Volatile Liquid

BACKGROUND

Most of the theory is covered in Book on Chemisty during courses Chemistry Course, Chem Engineering Course, and maybe Physics Course In the study of thermodynamics you have learnt or will be soon learning about the factors that determine when a pure liquid substance is in equilibrium with its vapor phase. Experimentally, for equilibrium to exist within a liquid-vapor system, it must be closed, that is the vapor is not disappearing into the ambient as it would if you had an open beaker of the liquid. Note that, in theory, we can choose an imaginary boundary encircling a smaller open system (where liquid and vapor exchange is allowed) within this bigger closed system.

At equilibrium the rate of vaporization of the liquid equals the rate of condensation of the vapor. One might imagine intuitively that these rates would be dependent upon the temperature of the system. Furthermore, the rate of condensation would also depend upon the concentration of the vapor, while the rate of evaporation should stay constant, because the density of a liquid remains constant at fixed temperature. From the ideal gas equation

at a specific temperature $T$, the gas concentration $n/V$ is proportional to its pressure $p$. Thus, it makes sense that at a given temperature there is a specific value for the liquid’s vapor pressure at equilibrium. We will explore this from a more quantitative, thermodynamic approach.

The condition at equilibrium is that the molar Gibbs free energy, or chemical potential of a pure liquid ${\mu }_L(T,p)$ is equal to that of the gas, ${\mu }_G(T,p)$. Recall that the differential for chemical potential is given by

where ${\stackrel{}{V¯}}_i$ and ${\stackrel{}{S¯}}_i$ are the molar volume and molar entropy, respectively, of the pure substances. As the chemical potential of the liquid and vapor are equal at equilibrium, so are the differentials for the chemical potential. Thus,

As we are interested in how vapor pressure changes with temperature, we solve for $\frac{\mathit{dp}}{\mathit{dT}}$:

where $\mathrm{\Delta }{\stackrel{}{S¯}}_{\mathit{vap}}$ and $\mathrm{\Delta }{\stackrel{}{V¯}}_{\mathit{vap}}$are the changes of molar entropy and molar volume associated with vaporization of our pure liquid. Again, recall that at equilibrium the chemical potential (i.e. molar free energy) of liquid and vapor are equal or, equivalently, the change of molar Gibbs free energy, $\mathrm{\Delta }{\stackrel{}{G¯}}_{\mathit{vap}}$, for liquid-vapor transition is zero. From the definition of Gibbs free energy we have

Substituting $\mathrm{\Delta }{\stackrel{}{S¯}}_{\mathit{vap}}$ into Equation (4), we have

Equations (4) and (6) are forms of the Clapeyron Equation.

Now to substitute for $\mathrm{\Delta }{\stackrel{}{V¯}}_{\mathit{vap}}$ we note that the molar volume of the vapor is much greater than that of the liquid, so we can safely approximate the change of molar volume by just the molar volume of the gas. We further approximate the Clapeyron Equation by replacing molar volume via the ideal gas law using Equation (1) and obtain

or, after rearranging,

Equation (8) is known as the Clausius-Clapeyron Equation.

By integrating this expression from some standard pressure $\mathit{p°}$ (where equilibrium is reached at boiling temperature $\mathit{T°}$) to actual pressure $p$ and temperature $T$ and assuming that $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}$ is independent of temperature over the range of vapor pressure measurements (i.e. $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}(T)=\mathrm{\Delta }\stackrel{}{H¯}{°}_{\mathit{vap}}$), we have

Observing that according to Equation (5), ($\mathrm{\Delta }\stackrel{}{H¯}{°}_{\mathit{vap}})/(\mathit{T°}=\mathrm{\Delta }{\stackrel{}{S¯}}_{\mathit{vap}}$) and setting $\mathit{p°}=1\mathrm{atm}$, we finally have:

The point of this derivation comes down to Equation (10). If one can measure the vapor pressure over a liquid as a function of temperature and one assumes that this vapor follows the ideal gas law fairly closely, then by plotting the natural log of pressure versus $1/T$, one should get a straight line. The slope of this line will yield a value for the standard molar enthalpy of vaporization, while its intercept yields the standard molar entropy of vaporization.

There are at least two possibly significant approximations made in this derivation. Gasses tend to deviate from ideal behavior near their critical temperature. A measure of this non-ideality is the gas compressibility factor, $Z$,

Using this expression to substitute for $\stackrel{}{V¯}$ in the Clapeyron, Equation (6), Equation (9) becomes

and the slope of this plot versus $1/T$, if linear, is $-\mathrm{\Delta }\stackrel{}{H¯}{°}_{\mathit{vap}}/\mathit{ZR}$. The second assumption to get a linear result from Equation (10) 10 or (12) was that $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}$ is fairly independent of temperature. In fact, this assumption also breaks down near critical temperature (where $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}$ tends to zero) and, generally, $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}$ decreases slightly with increasing $T$ at temperatures far from the critical point. Recall that the unique critical temperature and pressure for a specific pure substance indicate the temperature and pressure beyond which the liquid and vapor phases are no longer distinguishable. Both $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}$ and the compressibility factor $Z$ decrease with $T$ and, as a result, the ratio $\mathrm{\Delta }{\stackrel{}{H¯}}_{\mathit{vap}}/Z$ stays fairly constant.

In this experiment, we will be using a dynamic method to measure the vapor pressure of a pure liquid as a function of pressure. Recall that at the temperature at which a liquid begins to boil, the vapor pressure of that liquid is equal to the external pressure. In our apparatus, we will be controlling the pressure above our liquid sample by using lab air and a vacuum system. Maintaining the pressure to a near constant value while our sample liquid boils will be done by having a relatively large ballast volume as part of our system. Pressure will be measured by an electronic pressure gauge. The boiling liquid will be refluxed.