The Clapeyron equation for vaporization at low pressure is often simplified through reasonable approximations, namely, that the vapor phase is an ideal gas and that the molar volume of the liquid is negligible compared with the molar volume of the vapor.

How do these assumptions alter the Clapeyron equation?

Question

The Clapeyron equation for vaporization at low pressure is often simplified through reasonable approximations, namely, that the vapor phase is an ideal gas and that the molar volume of the liquid is negligible compared with the molar volume of the vapor.

How do these assumptions alter the Clapeyron equation?

Accepted Answer

For the assumptions made,

[latex] \Delta Z^{l v}=Z^v-Z^l=\frac{P^{s a t} V^v}{R T}-\frac{P^{s a t} V^l}{R T}=1-0=1 [/latex]

Then by Eq. (6.87),

[latex] \frac{d \ln P^{ ext {sat }}}{d T}=\frac{\Delta H^{l v}}{R T^2 \Delta Z^{l v}} [/latex] (6.87)

[latex] \Delta H^{l v}=-R \frac{d \ln P^{\mathrm{sat}}}{d(1 / T)} [/latex]

Known as the Clausius/Clapeyron equation, this approximate expression relates the latent heat of vaporization directly to the vapor-pressure curve. Specifically, it indicates a direct proportionality of [latex]ΔH^{lv} [/latex] to the slope of a plot of ln [latex]P^{sat} [/latex] vs. 1/T. Such plots of experimental data produce lines for many substances that are very nearly straight. The Clausius/ Clapeyron equation implies, in such cases, that [latex]ΔH^{lv} [/latex] is constant, virtually independent of T. This is not in accord with experiment; indeed, [latex]ΔH^{lv} [/latex] decreases monotonically with increasing temperature from the triple point to the critical point, where it becomes zero. The assumptions upon which the Clausius/ Clapeyron equation are based have approximate validity only at low pressures.

Question 6.7: The Clapeyron equation for vaporization at low pressure is o...

Related Answered Questions

Determine the vapor pressure (in kPa) for liquid n-hexane at 0, 30, 60, and 90°C: (a) With constants from App. B.2. (b) From the Lee/Kesler correlation for P^{r}_{sat}. ...

Verified Answer:

Estimate V, H^R, and S^R for an equimolar mixture of carbon dioxide(1) and propane(2) at 450 K and 140 bar by the Lee/Kesler correlations. ...

Verified Answer:

Supercritical CO2 can be used as an environmentally friendly solvent for cleaning applications, ranging from dry cleaning clothing to degreasing machine parts to photoresist stripping. A key advantage of CO2 is the ease with which it is separated from “dirt” and detergents. When its ...

Verified Answer:

Estimate V, U, H, and S for 1-butene vapor at 200°C and 70 bar if H and S are set equal to zero for saturated liquid at 0°C. Assume that the only data available are: Tc = 420.0 K Pc = 40.43 bar ω = 0.191 Tn = 266.9 K ( normal boiling point ) ...

Verified Answer:

A 1.5 m³ tank contains 500 kg of liquid water in equilibrium with pure water vapor, which fills the remainder of the tank. The temperature and pressure are 100°C and 101.33 kPa. From a water line at a constant temperature of 70°C and a constant pressure somewhat above 101.33 kPa, 750 kg of ...

Verified Answer:

A standard propane tank, like those typically used with a backyard gas grill, has an internal volume of 21.5 L. Using data from the NIST Chemistry WebBook, (a) find the mass and volume of liquid and vapor phase propane in such a tank containing a total of 20 lbm of propane at 25°C, and (b) find ...

Verified Answer:

Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and ΔH for P1 = 1000 kPa, t1 = 250°C, and P2 = 200 kPa. ...

Verified Answer:

Values of the second and third virial coefficients of nitrogen trifluoride are as follows: B = −94.3 cm^3·mol^−1 and C = 6740 cm^6.mol^−2 at 290 K B = −87.1 cm^3.mol^−1 and C = 6430 cm^6.mol^−2 at 300 K ...

Verified Answer:

Develop the property relations appropriate to the incompressible fluid, a model fluid for which both β and κ are zero (Sec. 3.2). This is an idealization employed in fluid mechanics. ...

Verified Answer:

Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information: 1. Table 6.1 gives compressibility-factor data (values of Z) for isobutane vapor. 2. The vapor pressure of isobutane at 360 K is 15.41 bar.3. Set H_{0)^{ig} = 18,115.0 J . mol^−1 and ...

Verified Answer: