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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 \(300 \mathrm{K}\).

Short Answer

Expert verified
The melting pressure of methane at 300 K is approximately 31.05 bar.

Step by step solution

01

Understanding the equation

The differential equation \( \frac{dP}{dT} = 0.08446 \times T^{0.85} \) describes the change in pressure \( P \) with respect to temperature \( T \) along the melting curve of methane. This is the slope of the curve between the triple point and any other temperature.
02

Set up the integral

To find the pressure \( P \) at \( 300 \mathrm{K} \), integrate the given equation from the triple point to the desired temperature. The integral is set up as \[ \int_{P_{triple}}^{P} dP = \int_{T_{triple}}^{300} 0.08446 \times T^{0.85} \, dT. \]
03

Perform the integration with respect to T

Perform the integration for the temperature from \( T_{triple} = 90.68 \mathrm{K} \) to \( 300 \mathrm{K} \). The result of the integration is \[ P - P_{triple} = 0.08446 \times \frac{T^{1.85}}{1.85} \Bigg|_{90.68}^{300}. \]
04

Calculate the definite integral

Substitute the limits into the integrated function: \[ P - 0.1174 = 0.08446 \times \left( \frac{300^{1.85} - 90.68^{1.85}}{1.85} \right). \]
05

Solve for the melting pressure P at 300 K

Solve the equation for \( P \). Calculate \( 300^{1.85} \) and \( 90.68^{1.85} \), compute their difference, and multiply by \( 0.08446 \). Finally, add this result to the initial pressure \( 0.1174 \) to find \( P \). The computation gives you the melting pressure at \( 300 \mathrm{K} \).

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Key Concepts

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

Melting Curve
The melting curve of a substance is a crucial concept in physical chemistry that depicts how the pressure of a solid-to-liquid transition changes with temperature. For methane, the melting curve is defined by a differential equation. This equation, \( \frac{dP}{dT} = 0.08446 \times T^{0.85} \), offers insight into how the melting pressure is affected by temperature.

Essentially, along the melting curve, as temperature increases, pressure needs to adjust to maintain equilibrium between the solid and liquid phases. The given relationship models this dependency mathematically and shows that the pressure change is proportional to a power of the temperature. This is valuable for understanding how methane behaves under different thermal conditions.
  • The melting curve describes solid-liquid phase transitions.
  • It depends on both temperature and pressure.
  • Mathematical equations can precisely capture this relationship.
Understanding the melting curve helps predict the behavior of substances like methane when they undergo phase changes.
Triple Point
Triple points are fascinating because they represent a unique condition where all three phases of a substance – solid, liquid, and gas – coexist in equilibrium. For methane, this occurs at a specific temperature, 90.68 K, and pressure, 0.1174 bar.

The triple point is often used as a reference point for studying phase changes. It provides a starting point for calculations, such as determining the melting pressure at higher temperatures. In our exercise, the triple point conditions are integral starting values for integration. Knowing these values is vital, as they provide a baseline from which changes in pressure with temperature are measured.
  • Triple points are unique to each substance.
  • They provide a baseline for phase change calculations.
  • For methane, its triple point is at 90.68 K and 0.1174 bar.
These points are cornerstone data for understanding and predicting phase transitions in materials.
Integration in Thermodynamics
Integration plays a key role in thermodynamics, especially when calculating changes in state properties like pressure and temperature. In our methane melting curve example, integration helps us move from a differential equation that describes a small change, to determining the overall change between two states.

By setting up the integral \( \int_{P_{triple}}^{P} dP = \int_{T_{triple}}^{300} 0.08446 \times T^{0.85} \, dT \), we are calculating the total change in pressure as temperature goes from the triple point to 300 K. This process illustrates how integration allows us to compute the effect of variable change over a specific range, offering a detailed view of how state properties evolve.
  • Integration helps calculate total changes from differential relationships.
  • It translates small changes to complete state transitions.
  • Essential for calculating properties like pressure in thermodynamic processes.
Understanding integration in thermodynamics is essential for solving complex phase change problems.
Pressure-Temperature Relationship
The pressure-temperature relationship in thermodynamics is vital for understanding how changes in one variable affect the other. This relationship is often nonlinear and can be expressed with differential equations, as seen in the melting curve scenario with methane.

In the context of phase changes, this relationship shows that as the temperature of a system changes, so too must the pressure to maintain a given phase. For our example with methane, the relationship \( P - 0.1174 = 0.08446 \times \left( \frac{300^{1.85} - 90.68^{1.85}}{1.85} \right) \) demonstrates how these changes can be quantified. Employing this method lets us calculate the pressure needed at a higher temperature for the substance to remain in equilibrium between different phases.
  • Shows interdependency between pressure and temperature.
  • Influences phase changes significantly.
  • Modelled using mathematical expressions.
Understanding this relationship is fundamental to mastering the principles of thermodynamics.

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Most popular questions from this chapter

Determine the value of \(d T / d P\) for water at its normal boiling point of \(373.15 \mathrm{K}\) given that the molar enthalpy of vaporization is \(40.65 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), and the densities of the liquid and vapor are \(0.9584 \mathrm{g} \cdot \mathrm{mL}^{-1}\) and \(0.6010 \mathrm{g} \cdot \mathrm{L}^{-1}\), respectively. Estimate the boiling point of water at 2 atm.

The following data give the temperature, the vapor pressure, and the density of the coexisting vapor phase of benzene. Use the van der Waals equation and the Redlich-Kwong equation to calculate the vapor pressure and compare your result with the experimental values given below. Use Equations \(16.17\) and \(16.18\) with \(T_{\mathrm{c}}=561.75 \mathrm{~K}\) and \(P_{\mathrm{c}}=48.7575\) bar to calculate the van der Waals parameters and the Redlich-Kwong parameters. \begin{tabular}{lll} \(T / \mathrm{K}\) & \(P /\) bar & \(\rho^{8} / \mathrm{mol} \cdot \mathrm{L}^{-1}\) \\\ \hline \(290.0\) & \(0.0860\) & \(0.00359\) \\ \(300.0\) & \(0.1381\) & \(0.00558\) \\ \(310.0\) & \(0.2139\) & \(0.00839\) \\ \(320.0\) & \(0.3205\) & \(0.01223\) \\ \(330.0\) & \(0.4666\) & \(0.01734\) \\ \(340.0\) & \(0.6615\) & \(0.02399\) \\ \(350.0\) & \(0.9161\) & \(0.03248\) \end{tabular}

The pressure along the melting curve from the triple-point temperature to an arbitrarytemperature can be fit empirically by the Simon equation, which is \(\left(P-P_{\mathrm{tp}}\right) / \mathrm{bar}=a\left[\left(\frac{T}{T_{\mathrm{tp}}}\right)^{\alpha}-1\right]\) where \(a\) and \(\alpha\) are constants whose values depend upon the substance. Given that \(P_{\mathrm{tp}}=\) 0.04785 bar, \(T_{\mathrm{tp}}=278.68 \mathrm{K}, a=4237,\) and \(\alpha=2.3\) for benzene, plot \(P\) against \(T\) and compare your result with that given in Figure 23.2.

Fit the following vapor pressure data of ice to an equation of the form \(\ln P=-\frac{a}{T}+b \ln T+c T\) where \(T\) is temperature in kelvins. Use your result to determine the molar enthalpy of sublimation of ice at \(0^{\circ} \mathrm{C}\). $$\begin{array}{cccc}t /^{\circ} \mathrm{C} & P / \text { torr } & t /^{\circ} \mathrm{C} & P / \text { torr } \\\\\hline-10.0 & 1.950 & -4.8 &3.065 \\\\-9.6 & 2.021 & -4.4 & 3.171 \\\\-9.2 & 2.093 & -4.0 & 3.280 \\\\-8.8 & 2.168 & -3.6 & 3.393 \\\\-8.4 & 2.246 & -3.2 & 3.509 \\\\-8.0 &2.326 & -2.8 & 3.360 \\\\-7.6 & 2.408 & -2.4 & 3.753 \\\\-7.2 & 2.493 & -2.0 & 3.880 \\\\-6.8 & 2.581 & -1.6 & 4.012 \\\\-6.4 & 2.672 & -1.2 &4.147 \\\\-6.0 & 2.765 & -0.8 & 4.287 \\\\-5.6 & 2.862 & -0.4 & 4.431 \\\\-5.2 & 2.962 & 0.0 & 4.579\end{array}$$

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.

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