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Chapter 23: Problem 23

The pressures at the solid-liquid coexistence boundary of propane are given by the empirical equation $$ P=-718+2.38565 T^{1.283} $$ where \(P\) is in bars and \(T\) is in kelvins. Given that \(T_{\text {fus }}=85.46 \mathrm{~K}\) and \(\Delta_{\text {fus }} \bar{H}=3.53 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), calculate \(\Delta_{\text {fus }} \bar{V}\) at \(85.46 \mathrm{~K}\).

Short Answer

Expert verified
\(\Delta_{\text{fus}} \bar{V}\) at 85.46 K is calculated using the Clapeyron equation after differentiating the pressure equation and converting units.

Step by step solution

01

Identify the Given Values

We are provided with the following values:- The empirical equation for pressure: \(P = -718 + 2.38565 \times T^{1.283}\)- The fusion temperature: \(T_{\text{fus}} = 85.46 \, \text{K}\)- The change in enthalpy of fusion: \(\Delta_{\text{fus}} \bar{H} = 3.53 \, \text{kJ} \cdot \text{mol}^{-1}\).
02

Calculate Pressure at Fusion Temperature

Substitute the value of \(T_{\text{fus}} = 85.46 \, \text{K}\) into the pressure equation:\[ P = -718 + 2.38565 \times (85.46)^{1.283} \]Calculate \((85.46)^{1.283}\) and then find \(P\).
03

Use Clapeyron Equation

The Clapeyron equation relates the change in pressure to the change in volume and enthalpy:\[ \frac{dP}{dT} = \frac{\Delta_{\text{fus}} \bar{H}}{T \Delta_{\text{fus}} \bar{V}} \]We need \(\Delta_{\text{fus}} \bar{V}\), so rearrange the equation to:\[ \Delta_{\text{fus}} \bar{V} = \frac{\Delta_{\text{fus}} \bar{H}}{T \cdot \frac{dP}{dT}} \].
04

Differentiate Pressure Equation

Differentiate the pressure equation with respect to temperature, \(T\):\[ \frac{dP}{dT} = 2.38565 \times 1.283 \times T^{0.283} \].
05

Calculate Derivative at Fusion Temperature

Substitute \(T = 85.46\) into the differentiated equation to find \(\frac{dP}{dT}\):\[ \frac{dP}{dT} = 2.38565 \times 1.283 \times (85.46)^{0.283} \]Calculate \((85.46)^{0.283}\) and then find \(\frac{dP}{dT}\).
06

Compute Change in Volume \(\Delta_{\text{fus}} \bar{V}\)

Use the Clapeyron equation to compute \(\Delta_{\text{fus}} \bar{V}\):\[ \Delta_{\text{fus}} \bar{V} = \frac{3.53 \, \text{kJ} \cdot \text{mol}^{-1}}{85.46 \, \text{K} \cdot \frac{dP}{dT}} \]Ensure \(\Delta_{\text{fus}} \bar{H}\) is in the same units as pressure by converting from kJ to J (1 kJ = 1000 J).
07

Perform Calculations and Units Conversion

Convert \(\Delta_{\text{fus}} \bar{H}\) from kJ/mol to J/mol:\[ \Delta_{\text{fus}} \bar{H} = 3.53 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 3530 \, \text{J/mol} \]Substitute into the equation for \(\Delta_{\text{fus}} \bar{V}\) and compute the final value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Fusion
Enthalpy of fusion is the energy required to change a substance from solid to liquid at constant pressure and temperature. This is an important concept in understanding phase changes.
  • It reflects how much energy is needed to overcome intermolecular forces.
  • In the context of propane, the given enthalpy of fusion is 3.53 kJ/mol.
  • This means that 3.53 kJ of energy is required for each mole of solid propane to become liquid at the fusion temperature of 85.46 K.
Enthalpy of fusion is usually expressed in kilojoules per mole (kJ/mol). In calculations, proper unit conversion is crucial (e.g., converting from kJ to J for consistency in equations). This parameter also affects the Clapeyron equation, which is used to model phase equilibrium while taking enthalpy, pressure, and volume into account.
Solid-Liquid Equilibrium
When a substance is in solid-liquid equilibrium, the solid and liquid phases coexist at a specific temperature and pressure. For propane, this equilibrium occurs at its fusion temperature of 85.46 K.
  • The Clapeyron equation is used to describe the conditions where solid and liquid phases are in balance.
  • At equilibrium, the molecular exchange between phases is steady, allowing the phases to coexist.
  • Understanding this balance is crucial for thermodynamic calculations, as it provides insight into melting and freezing processes.
Equilibrium points are essential in understanding the material properties under different conditions. They enable prediction of how a substance will behave given changes in temperature and pressure, helping to establish conditions for controlled changes like melting.
Differentiation of Equations
Differentiation is a mathematical process used to determine how a function changes at any point. In the context of the Clapeyron equation, differentiating the given pressure equation enables us to ascertain how pressure changes with temperature.
  • The differentiation helps find the rate of change of pressure with respect to temperature, i.e., the derivative \(\frac{dP}{dT}\).
  • This derivative is a critical component of the Clapeyron equation, used to calculate changes in volume during fusion.
  • The calculation involves applying the power rule and chain rule for differentiation to the given empirical equation for pressure.
Differentiation forms the foundation for solving problems related to dynamic changes in systems. It allows the transformation of complex empirical relationships into usable data for predicting behaviors of substances under varying conditions.
Pressure-Temperature Relationship
The pressure-temperature relationship is fundamental to understanding how different states of matter interact and transition. It's described empirically by equations like the one given for propane:
  • This relationship indicates how pressure changes in response to temperature around the solid-liquid boundary.
  • For this exercise, we're looking at the impact of temperature changes on pressure using an exponential model: \(P = -718 + 2.38565 T^{1.283}\).
  • The relationship is critical for calculating the volume change upon fusion through the Clapeyron equation.
Understanding this relationship helps predict how propane鈥攍ike other substances鈥攚ill behave near its melting point. It serves as a foundational principle for more complex analyses in thermodynamics, chemical engineering, and physical chemistry.

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Most popular questions from this chapter

The vapor pressure of mercury from \(400^{\circ} \mathrm{C}\) to \(1300^{\circ} \mathrm{C}\) can be expressed by \(\ln (P / \text { torr })=-\frac{7060.7 \mathrm{K}}{T}+17.85\) The density of the vapor at its normal boiling point is \(3.82 \mathrm{g} \cdot \mathrm{L}^{-1}\) and that of the liquid is \(12.7 \mathrm{g} \cdot \mathrm{mL}^{-1} .\) Estimate the molar enthalpy of vaporization of mercury at its normal boiling point.

The slope of the melting curve of methane is given by \(\frac{d P}{d T}=\left(0.08446 \mathrm{bar} \cdot \mathrm{K}^{-1.85}\right) T^{0.85}\) from the triple point to arbitrary temperatures. Using the fact that the temperature and pressure at the triple point are \(90.68 \mathrm{K}\) and 0.1174 bar, calculate the melting pressure of methane at \(300 \mathrm{K}\).

The densities of the coexisting liquid and vapor phases of methanol from the triple point to the critical point are accurately given by the empirical expressions \(\frac{\rho^{1}}{\rho_{\mathrm{c}}}-1=2.51709(1-x)^{0.350}+2.466694(1-x)\) \(-3.066818\left(1-x^{2}\right)+1.325077\left(1-x^{3}\right)\) and \(\ln \frac{\rho^{\mathrm{g}}}{\rho_{\mathrm{c}}}=-10.619689 \frac{1-x}{x}-2.556682(1-x)^{0.350}\) \(+3.881454(1-x)+4.795568(1-x)^{2}\) where \(\rho_{\mathrm{c}}=8.40 \mathrm{mol} \cdot \mathrm{L}^{-1}\) and \(x=T / T_{\mathrm{c}},\) where \(T_{\mathrm{c}}=512.60 \mathrm{K} .\) Use these expressions to plot \(\rho^{\prime}\) and \(\rho^{8}\) against temperature, as in Figure \(23.7 .\) Now plot \(\left(\rho^{1}+\rho^{8}\right) / 2\) against \(T\). Show that this line intersects the \(\rho^{1}\) and \(\rho^{\mathrm{g}}\) curves at \(T=T_{\mathrm{c}}\).

Use the following data to estimate the normal boiling point and the molar enthalpy of vaporization of lead. $$\begin{array}{l|lllll} T / \mathrm{K} & 1500 & 1600 & 1700 & 1800 & 1900 \\ \hline P / \text { torr } & 19.72 & 48.48 & 107.2 & 217.7 & 408.2 \end{array}$$

The orthobaric densities of liquid and gaseous ethyl acetate are \(0.826 \mathrm{g} \cdot \mathrm{mL}^{-1}\) and \(0.00319 \mathrm{g} \cdot \mathrm{mL}^{-1},\) respectively, at its normal boiling point \(\left(77.11^{\circ} \mathrm{C}\right) .\) The rate of change of vapor pressure with temperature is 23.0 torr \(\cdot \mathrm{K}^{-1}\) at the normal boiling point. Estimate the molar enthalpy of vaporization of ethyl acetate at its normal boiling point.
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