Question

In Equations \((8.16),(d P / d T)_{\text {vaporization}}\) was calculated by assuming that \(V_{m}^{g a s} \gg V_{m}^{\text {liquid}} .\) In this problem, you will test the validity of this approximation. For water at its normal boiling point of \(373.12 \mathrm{K}, \Delta H_{\text {vaporization}}=40.656 \times\) \(10^{3} \mathrm{Jmol}^{-1}, \rho_{\text {liquid}}=958.66 \mathrm{kg} \mathrm{m}^{-3},\) and \(\rho_{\text {gas}}=\) \(0.58958 \mathrm{kg} \mathrm{m}^{-3} .\) Compare the calculated values for \((d P / d T)_{\text {vaporization}}\) with and without the approximation of Equation \((8.16) .\) What is the relative error in making the approximation?

Short answer

The molar volumes for the liquid and gas are calculated as: \(V_m^{liquid} = \frac{18.015 \times 10^{-3} \mathrm{kg/mol}}{958.66 \mathrm{kg/m^3}} \approx 1.88 \times 10^{-5} \mathrm{m^3/mol}\) \(V_m^{gas} = \frac{18.015 \times 10^{-3} \mathrm{kg/mol}}{0.58958 \mathrm{kg/m^3}} \approx 3.05 \times 10^{-2} \mathrm{m^3/mol}\) Using these molar volumes in the Clapeyron equation, we find: \(\frac{dP}{dT} = \frac{40.656 \times 10^3 \mathrm{J/mol}}{373.12 \mathrm{K} ((3.05 \times 10^{-2}) - (1.88 \times 10^{-5}))\mathrm{m^3/mol}} \approx 1167.34 \mathrm{Pa/K}\) (without approximation) \(\frac{dP}{dT} \approx \frac{40.656 \times 10^3 \mathrm{J/mol}}{373.12 \mathrm{K} (3.05 \times 10^{-2} \mathrm{m^3/mol})} \approx 1168.15 \mathrm{Pa/K}\) (with approximation) The relative error in making the approximation is: \(Relative\ Error = \frac{|1167.34 - 1168.15|}{1167.34} \approx 0.00069\) Hence, the relative error in making the approximation of \(V_m^{gas} \gg V_m^{liquid}\) is approximately 0.069%.

Step-by-step solution

Find the Molar Volume

Molar volume is defined as the volume occupied by 1 mole of a substance, and is related to density by: \[V_m = \frac{M}{\rho}\] where M is the molar mass of the substance. For the liquid, molar volume (\(V_m^{liquid}\)) can be calculated using the density (\(\rho_{liquid}\)) and the molar mass of water (M = 18.015 g/mol): \[V_m^{liquid} = \frac{18.015 \times 10^{-3} \mathrm{kg/mol}}{958.66 \mathrm{kg/m^3}}\] Similarly, for the gas, molar volume (\(V_m^{gas}\)) can be calculated using the density (\(\rho_{gas}\)) and the molar mass of water (M = 18.015 g/mol): \[V_m^{gas} = \frac{18.015 \times 10^{-3} \mathrm{kg/mol}}{0.58958 \mathrm{kg/m^3}}\]

Calculate \((dP/dT)_{vaporization}\) without approximation

Clapeyron equation is given by: \[\frac{dP}{dT} = \frac{\Delta H_{vaporization}}{T (V_m^{gas} - V_m^{liquid})}\] Now plug in the values we calculated for the molar volumes, \(\Delta H_{vaporization}\), and the temperature (T = 373.12 K): \[\frac{dP}{dT} = \frac{40.656 \times 10^3 \mathrm{J/mol}}{373.12 \mathrm{K} ((V_m^{gas}) - (V_m^{liquid}))}\]

Calculate \((dP/dT)_{vaporization}\) with approximation

Considering the approximation that \(V_m^{gas} \gg V_m^{liquid}\) in the Clapeyron equation, we get: \[\frac{dP}{dT} \approx \frac{\Delta H_{vaporization}}{T (V_m^{gas})}\] Again, plug in the values for \(\Delta H_{vaporization}\), T = 373.12 K, and \(V_m^{gas}\) to calculate \(\frac{dP}{dT}\): \[\frac{dP}{dT} \approx \frac{40.656 \times 10^3 \mathrm{J/mol}}{373.12 \mathrm{K} (V_m^{gas})}\]

Calculate the relative error

First, let's denote the exact value of \(\frac{dP}{dT}\) as \(dPdT_{exact}\), and the approximate value as \(dPdT_{approx}\). The relative error can be calculated as follows: \[Relative\ Error = \frac{|dPdT_{exact} - dPdT_{approx}|}{dPdT_{exact}}\] Use the values obtained in step 2 and step 3 to find the relative error in making the approximation.

Key concepts

Vaporization

Vaporization is a phase transition process in which a substance changes from its liquid phase to a gaseous phase. It is an essential concept in thermodynamics, particularly when discussing boiling and evaporation processes. During vaporization, energy in the form of heat (enthalpy of vaporization, \(\Delta H_{\text{vaporization}}\)) is absorbed by the liquid, overcoming the intermolecular forces and allowing the molecules to spread out more freely and become a gas. Understanding the energetics and dynamics of vaporization is crucial for various applications ranging from industrial chemical processes to predicting weather patterns.

When considering vaporization in thermodynamic calculations, it is often assumed that the molar volume of the gas, \(V_m^{gas}\), is much greater than the molar volume of the liquid, \(V_m^{liquid}\). This simplification can make mathematical modeling more straightforward. However, it's important to assess the validity of such an assumption, as it can introduce errors into the calculations, leading to imprecise understanding of the behavior of a substance as it transitions from liquid to gas.

Molar Volume

Molar volume, \(V_m\), is the volume occupied by one mole of a substance, and it is a critical measurement in understanding the properties of materials in different phases. It is particularly significant in the study of gases, where it helps in the application of the ideal gas law in conditions where gases behave predictably. The molar volume is calculated by dividing the molar mass of a substance, \(M\), by its density, \(\rho\).

For the case of water vaporization, the molar volume for both the liquid and the gas phase can be obtained by using their respective densities. As we see from the solution, these values have a huge discrepancy, indicating the substantially larger volume occupied by water in the gaseous state compared to the liquid state. This highlights the general concept that gases occupy far more volume than liquids for the same number of moles, a fact that is foundational in the sciences of chemistry and physics.

Phase Transition

A phase transition involves a substance changing from one state of matter to another. Classic examples are transitions between solid, liquid, and gas phases. Each phase transition—like melting, freezing, vaporization, condensation, sublimation, and deposition—occurs at specific temperature and pressure conditions and often involves the absorption or release of heat. During a phase transition, the temperature of a system remains constant while the substance absorbs or releases heat energy until the transition is complete.

Understanding phase transitions is important for analyzing the behavior of materials under various environmental conditions. The Clapeyron equation, which connects the pressure and temperature during a phase transition, provides crucial insights into such processes. It also allows the calculation of the enthalpy change (\(\Delta H\)) associated with the transition, such as the amount absorbed during vaporization. In calculations, ensuring that the correct physical properties of the substances involved are used is paramount to achieve accurate results.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. The field plays a critical role in engineering, chemistry, and biology, providing key principles that govern everything from engines to ecosystems. Central to thermodynamics are the laws that describe the conservation of energy, the increase of entropy, and the concept of equilibrium.

Applying thermodynamics concepts, such as the Clapeyron equation, allows for the precise calculation of temperature and pressure-related changes during phase transitions like vaporization. The equation shows the proportional relationship between the change in pressure over the change in temperature and the entropy change, involving molar volumes and vaporization enthalpies. As such, understanding basic thermodynamic principles is essential for interpreting phase transitions and predicting the behavior of substances under various conditions.