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Chapter 11: Problem 9

When lead is melted at atmospheric pressure, the melting point is \(600 \mathrm{~K}\), the density decreases from \(11.01\) to \(10.65 \mathrm{~g} / \mathrm{cm}^{3}\), and the latent heat of fusion \(\Delta H\) is \(24.5 \mathrm{~J} / \mathrm{g}\). What is the melting point at the pressure of \(1.01 \times 10^{7} \mathrm{~Pa}\) ?

Short Answer

Expert verified
The melting point at the pressure of \(1.01 \times 10^7 \) Pa is 607.6 K.

Step by step solution

01

- Understand the Given Data

Identify the given data: 1. Melting point at atmospheric pressure: \( T_m = 600 \text{ K} \) 2. Initial density (solid lead): \( \rho_s = 11.01 \text{ g/cm}^3 \) 3. Final density (liquid lead): \( \rho_l = 10.65 \text{ g/cm}^3 \) 4. Latent heat of fusion: \( \triangle H = 24.5 \text{ J/g} \) 5. Desired pressure: \( P_2 = 1.01 \times 10^7 \text{ Pa} \)
02

- Convert Densities to Consistent Units

Convert densities from g/cm^3 to kg/m^3: \[ \rho_s = 11.01 \text{ g/cm}^3 = 11010 \text{ kg/m}^3 \] \[ \rho_l = 10.65 \text{ g/cm}^3 = 10650 \text{ kg/m}^3 \]
03

- Calculate Change in Volume

Calculate the change in specific volume \( \triangle v \): \[ \triangle v = v_l - v_s = \frac{1}{\rho_l} - \frac{1}{\rho_s} \] \[ \triangle v = \frac{1}{10650} - \frac{1}{11010} \text{ m}^3/\text{kg} \] \[ \triangle v = 9.39 \times 10^{-8} - 9.08 \times 10^{-8} \text{ m}^3/\text{kg} \] \[ \triangle v = 3.1 \times 10^{-9} \text{ m}^3/\text{kg} \]
04

- Apply the Clausius-Clapeyron Equation

Use the Clausius-Clapeyron equation to find the melting point at the new pressure: \[ \frac{dT_m}{dP} = \frac{T_m \triangle v}{\triangle H} \] Rearrange to find the change in temperature: \[ \triangle T = \frac{T_m \triangle v \triangle P}{\triangle H} \] Given \( \triangle P = P_2 - P_1 \) and \( P_1 = 1.01 \times 10^5 \text{ Pa} \)
05

- Calculate the Change in Temperature

Calculate the change in temperature \( \triangle T \): \[ \triangle T = \frac{600 \text{ K} \times 3.1 \times 10^{-9} \text{ m}^3/\text{kg} \times (1.01 \times 10^7 - 1.01 \times 10^5) \text{ Pa}}{24.5 \text{ J/g} \times \frac{1 \text{ kg}}{1000 \text{ g}}} \] \[ \triangle T = \frac{600 \text{ K} \times 3.1 \times 10^{-9} \times 1.0 \times 10^7}{24.5 \times 10^{-3}} \] \[ \triangle T = \frac{600 \times 3.1 \times 10^{-9} \times 10^7}{24.5 \times 10^{-3}} \] \[ \triangle T = \frac{600 \times 3.1 \times 10^{-2}}{24.5 \times 10^{-3}} \] \[ \triangle T \text{ = 7.6 K} \]
06

- Calculate the New Melting Point

Add the change in temperature to the initial melting point to find the new melting point: \[ T_{m,new} = T_m + \triangle T \] \[ T_{m,new} = 600 \text{ K} + 7.6 \text{ K} \] \[ T_{m,new} = 607.6 \text{ K} \]

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Key Concepts

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

Melting Point
The melting point is the temperature at which a solid substance transforms into a liquid. For lead at atmospheric pressure, this point is 600 K. However, the melting point can change when the pressure changes. This is described by the Clausius-Clapeyron equation, which allows us to predict how the melting point will shift with pressure variations. By increasing the pressure, we can raise the melting point of lead. In this exercise, when pressure is raised to 1.01 x 10^7 Pa, the melting point rises from 600 K to 607.6 K.
Specific Volume
Specific volume refers to the volume occupied by a unit mass of a substance. It is the inverse of density. When lead melts, its density changes from 11.01 g/cm^3 in its solid state to 10.65 g/cm^3 in its liquid state. To find the specific volume difference (\triangle v), we need to convert these densities to consistent units (kg/m^3) and then calculate: ewline ewline \[ \triangle v = \frac{1}{\rho_l} - \frac{1}{\rho_s} = \frac{1}{10650} - \frac{1}{11010} = 3.1 \times 10^{-9} \text{ m}^3/\text{kg} \]. Understanding this change is crucial as it influences how substances behave under different conditions and is vital for calculating the pressure effect on the melting point.
Latent Heat of Fusion
Latent heat of fusion (\triangle H) is the energy required to change a substance from its solid phase to its liquid phase at its melting point, without changing its temperature. For lead, this value is 24.5 J/g. This energy is necessary to overcome the forces holding the particles together in the solid state. Latent heat plays a key role in determining how much the melting point changes with pressure, as calculated using the Clausius-Clapeyron equation: ewline ewline \[ \frac{dT_m}{dP} = \frac{T_m \triangle v}{\triangle H} \]. By understanding this energy requirement, we can assess how changing conditions affect the phase transition.
Pressure Change
Pressure change is a crucial factor in modifying the melting point of substances. In this exercise, the initial pressure is atmospheric pressure (1.01 x 10^5 Pa), and the desired pressure is significantly higher (1.01 x 10^7 Pa). The Clausius-Clapeyron equation helps calculate the temperature change due to this pressure difference: ewline ewline \[ \triangle T = \frac{T_m \triangle v \triangle P}{\triangle H} = \frac{600 K \times 3.1 \times 10^{-9} \text{ m}^3/\text{kg} \times (1.01 \times 10^7 - 1.01 \times 10^5) \text{ Pa}}{24.5 \text{ J/g} \times \frac{1 \text{ kg}}{1000 \text{ g}}} = 7.6 K \]. Hence, the new melting point becomes 607.6 K.

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Most popular questions from this chapter

The latent heat of fusion for ice \(I\) is \(3.34 \times 10^{5} \mathrm{~J} / \mathrm{kg}\) at \(0^{\circ} \mathrm{C}\) and atmospheric pressure. If the change in specific volume on melting is \(-9.05 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{kg}\), then calculate the change of melting temperature due to change of pressure.

The Joule-Thomson coefficient \(\mu\) is a measure of the temperature change during a throttling process. A similar measure of the temperature change produced by an isentropic change of pressure is provided by the coefficient \(\mu_{s}\), where $$ \mu_{S}=\left(\frac{\partial T}{\partial P}\right)_{s} $$ Prove that $$ \mu_{S}-\mu=\frac{V}{C_{P}} $$

(a) Show that in a Joule-Thomson expansion, no temperature change occurs if \((\partial \nu / \partial T)_{P}=v / T\) (b) Show that $$ \mu c_{P}=T\left(\frac{\partial \nu}{\partial T}\right)_{P}-\nu . $$ In the region of moderate pressures, the equation of state of \(1 \mathrm{~mol}\) of gas may be written $$ P v=R T+B^{\prime} P+C^{\prime} P^{2} $$ where the second and third virial coefficients \(B^{\prime}\) and \(C^{\prime}\) are functions of \(T\) only. (c) Show that, as the pressure approaches zero, $$ \mu c_{P} \rightarrow T \frac{d B^{\prime}}{d T}-B^{\prime} $$ (d) Show that the equation of the inversion curve is $$ P=-\frac{B^{\prime}-T\left(d B^{\prime} / d T\right)}{C^{\prime}-T\left(d C^{\prime} / d T\right)} $$

Water at its freezing point \(\left(T_{i}, P_{i}\right)\) completely fills a strong steel container. The temperature is reduced to \(T_{f}\) at constant volume, with the pressure rising to \(P_{f}\). (a) Show that the fraction \(y\) of water that freezes is given by $$ y=\frac{v_{f}^{\prime \prime}-v_{f}^{n}}{v_{f}^{\prime \prime}-v_{f}^{\prime}} $$ (b) State explicitly the simplifying assumptions that must be made in order that \(y\) may be written $$ y=\frac{v^{\prime \prime}\left[\beta^{\prime \prime}\left(T_{f}-T_{i}\right)-\kappa^{\prime \prime}\left(P_{f}-P_{i}\right)\right]}{v_{f}^{\theta}-v_{f}^{\prime}} $$ (c) Calculate \(y\) for \(i=0^{\circ} \mathrm{C}, \quad 1.01 \times 10^{5} \mathrm{~Pa} ; \quad f=-5^{\circ} \mathrm{C}, \quad 5.98 \times 10^{7} \mathrm{~Pa}\); \(\beta^{\prime \prime}=-67 \times 10^{-6} \mathrm{~K}^{-1} ; \kappa^{\prime \prime}=12.04 \times 10^{-11} \mathrm{~Pa}^{-1} ; v_{f}^{\prime \prime}-v_{f}^{\prime}=-1.02 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{kg}\)

(a) Prove that, for a single phase, $$ \left(\frac{\partial P}{\partial T}\right)_{S}=\frac{c_{P}}{T u \beta} $$ (b) Calculate \((\partial P / \partial T)_{S}\) for ice at \(-3^{\circ} \mathrm{C}\), where \(c_{P}=2.01 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, v=1.09 \times\) \(10^{-3} \mathrm{~m}^{3} / \mathrm{kg}\), and \(\beta=1.58 \times 10^{-4} \mathrm{~K}^{-1}\) (c) Ice is initially at \(-3^{\circ} \mathrm{C}\) and \(1.01 \times 10^{5} \mathrm{~Pa}\). The pressure is increased adiabatically until the ice reaches the melting point. At what temperature and pressure is this melting point? (Hint: At what point does a line whose slope is \((\partial P / \partial T)_{S}\) cut a line whose slope is that of the fusion curve, \(\left.-1.35 \times 10^{7} \mathrm{~Pa} / \mathrm{K} ?\right)\)
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