Suppose that \(S\) is regarded as a function of \(p\) and \(\mathrm{T}\). Show that \(T \mathrm{dS}=\) \(C_{\mathrm{p}} \mathrm{d} T-\alpha T V \mathrm{~d} p .\) Hence, show that the energy transferred as heat when the pressure on an incompressible liquid or solid is increased by \(\Delta p\) is equal to \(-\alpha T V \Delta p .\) Evaluate \(q\) when the pressure acting on \(100 \mathrm{~cm}^{3}\) of mercury at \(0^{\circ} \mathrm{C}\) is increased by \(1.0\) kbar. \(\left(\alpha=1.82 \times 10^{-4} \mathrm{~K}^{-1} .\right)\)

Short Answer

Expert verified
The energy transferred as heat when the pressure on 100 cm\(^3\) of mercury at 0\(^\circ\)C is increased by 1.0 kbar is \(q = -4.977 \times 10^2 J\).

Step by step solution

01

Understanding Given Relation

The provided equation, which is a thermodynamic identity, relates the change in entropy (dS) to the change in temperature (dT) and pressure (dp). This equation is derived from the fundamental equation for the changes in heat in a system. It is given as: \(T \text{dS} = C_p \text{dT} - \alpha T V \text{dp}\) where \(C_p\) is the heat capacity at constant pressure, \(T\) is the temperature, \(V\) is the volume, and \(\alpha\) is the coefficient of thermal expansion.
02

Analyzing the Situation for an Incompressible Substance

An incompressible substance has a volume that does not change with pressure; therefore, its coefficient of volume expansion \(\beta\) is zero. Since \(\alpha = \frac{\beta}{K_T}\) where \(K_T\) is the isothermal compressibility and \(\beta\) is effectively zero, the term involving \(\alpha\), which relates to changes in volume with temperature, becomes significant in determining the heat transfer when pressure is changed.
03

Calculating Heat Transfer for Pressure Increase

Given that the substance is incompressible, there are no changes in temperature and volume when pressure changes, meaning that dT = 0 and dV = 0 in this case. Thus, the heat transferred can be simplified to just the work done by pressure, which is the second term in the identity: \(\delta q = - \alpha T V \Delta p \).
04

Determining Heat Transfer in Specific Conditions

Substitute given values into the simplified equation for heat transfer to find the heat q involved when the pressure on an incompressible substance is increased. Using the given coefficient of thermal expansion \(\alpha\), temperature at 0\(^\circ\)C (273.15K), volume of mercury (100 cm\(^3\)), and the increase in pressure (1.0 kbar which is \(1 \times 10^8\) Pa), calculate q.
05

Performing the Calculation

Using the values \(\alpha = 1.82 \times 10^{-4} K^{-1}\), \(T = 273.15 K\), \(V = 100 \times 10^{-6} m^3\) (since 1 cm\(^3\) is \(1 \times 10^{-6} m^3\)), and \(\Delta p = 1.0 \times 10^8 Pa\), plug these into the equation to compute the energy transferred as heat \(q\): \(q = - \alpha T V \Delta p = - (1.82 \times 10^{-4} K^{-1}) \times (273.15 K) \times (100 \times 10^{-6} m^3) \times (1.0 \times 10^8 Pa)\). Multiply these values to get the final result.

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Key Concepts

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

Entropy Change
Entropy is fundamentally understood as a measure of disorder or randomness in a system. Entropy change, often denoted as \( \Delta S \), refers to the difference in the level of disorder between two states of a system.

In thermodynamics, the change in entropy is closely related to the heat transfer within a system. At a particle level, when a system receives or releases heat, the movement and energy distribution among the particles are altered, which in turn affects its entropy.

The relationship between entropy change and heat transfer is especially important when examining reversible processes, where \( \Delta S = \int \frac{dq_{\text{rev}}}{T} \). For processes under constant pressure, the change in entropy can be represented using the heat capacity at constant pressure \( C_p \), as in the provided thermodynamic identity \( TdS = C_p dT - \alpha T V dp \). This means that when temperature or pressure changes, the entropy of the system changes in proportion to these variations.
Heat Capacity at Constant Pressure
The heat capacity at constant pressure, denoted as \( C_p \), is a quantity that describes how much heat energy is required to raise the temperature of a substance by a certain amount under constant pressure conditions.

It's an essential aspect of energy transfer within materials, as it reflects how substances respond to temperature changes. For instance, a high \( C_p \) indicates that a substance can absorb a lot of heat without a significant change in temperature, while a low \( C_p \) means the opposite. In the context of the aforementioned thermodynamic identity, \( C_p \) is the coefficient that scales the rise in temperature to the corresponding entropy change when pressure remains constant.

Engineers and scientists consider \( C_p \) crucial when designing systems where temperature regulation and heat exchange are involved, such as in cooling technologies or thermal insulation materials.
Coefficient of Thermal Expansion
The coefficient of thermal expansion, typically represented by the symbol \( \alpha \), quantifies how the size of a material changes with temperature. Specifically, it describes the fractional change in volume per degree of temperature change.

It's a critical parameter for materials that undergo thermal stresses, as structural components expand and contract with temperature fluctuations – for instance, in bridges or railroad tracks.

In our thermodynamic identity, the term \( -\alpha T V dp \) is associated with the work done against external pressure due to thermal expansion. It implies that even when a substance is incompressible, there can still be work interaction (and hence heat transfer) as a result of pressure changes. Materials with higher values of \( \alpha \) expand more for given temperature increases, influencing not only structural designs, but also the calculation of heat transfers and work in various thermodynamic processes.
Incompressible Substance Thermodynamics
When discussing the thermodynamics of incompressible substances, it's crucial to recognize materials such as liquids and solids where volume remains relatively constant under pressure changes. These materials are treated as having negligible compressibility.

In practical terms, for these substances, the association between temperature, pressure, and volume can be simplified since the volume change associated with pressure \( (\Delta V) \) is essentially zero. This simplification is used in the thermodynamic identity \( TdS = C_p dT - \alpha T V dp \), where the term involving \( dp \) (pressure change) becomes negligible in incompressible substances.

Consequently, when analyzing the effects of pressure on an incompressible liquid or solid, such as mercury in the given exercise, the heat transfer becomes primarily dependent on the material's coefficient of thermal expansion (\( \alpha \)) and the amount of pressure applied, rather than volume changes. The result is a direct and predictable interaction between pressure changes and heat transfer which is central to the design of processes involving such materials.

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Most popular questions from this chapter

The adiabatic compressibility, \(\kappa_{S}\), is defined like \(\kappa_{T}\) (eqn 2.44) but at constant entropy. Show that for a perfect gas \(p \gamma \kappa_{\mathrm{S}}=1\) (where \(\gamma\) is the ratio of heat capacities).

To calculate the work required to lower the temperature of an object, we need to consider how the coefficient of performance changes with the temperature of the object. (a) Find an expression for the work of cooling an object from \(\mathrm{T}\), to \(\mathrm{T}_{\mathrm{r}}\) when the refrigerator is in a room at a temperature \(\mathrm{T}_{\mathrm{h}}\). Hint. Write \(d w=d y l c\left(T\right.\) relate \(d q\) to \(\mathrm{dT}\) through the heat capacity \(\mathrm{C}_{p}\), and integrate the resulting expression. Assume that the heat capacity is independent of temperature in the range of interest. (b) Use the result in part (a) to calculate the work needed to freeze \(250 \mathrm{~g}\) of water in a refrigerator at \(293 \mathrm{~K}\). How long will it take when the refrigerator operates at \(100 \mathrm{~W}\) ?

The protein lysozyme unfolds at a transition temperature of \(75.5^{\circ} \mathrm{C}\) and the standard enthalpy of transition is \(509 \mathrm{k}) \mathrm{mol}^{-1} .\) Calculate the entropy of unfolding of lysozyme at \(25.0^{\circ} \mathrm{C}\), given that the difference in the constantpressure heat capacities upon unfolding is \(6.28 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and can be assumed to be independent of temperature. Hint. Imagine that the transition at \(25.0^{\circ} \mathrm{C}\) occurs in three steps: (i) heating of the folded protein from \(25.0^{\circ} \mathrm{C}\) to the transition temperature, (ii) unfolding at the transition temperature, and function, the entropy change at \(25.0^{\circ} \mathrm{C}\) is equal to the sum of the entropy changes of the steps. (iii) cooling of the unfolded protein to \(25.0^{\circ} \mathrm{C}\). Because the entropy is a state

Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature \(T_{\mathrm{h}}\) and the other at \(T_{c}\), are brought into thermal contact and allowed to reach equilibrium. Evaluate the change for two blocks of copper, each of mass \(500 \mathrm{~g}\), with \(C_{p, m}=24.4 \mathrm{~J} \mathrm{~K}^{-1}\) \(\mathrm{mol}^{-1}\), taking \(T_{\mathrm{h}}=500 \mathrm{~K}\) and \(T_{\mathrm{c}}=250 \mathrm{~K}\)

A gaseous sample consisting of \(1.00 \mathrm{~mol}\) molecules is described by the equation of state \(p \mathrm{~V}_{\mathrm{m}}=\mathrm{RT}(1+B p)\). Initially at \(373 \mathrm{~K}\), it undergoes JouleThomson expansion from \(100 \mathrm{~atm}\) to \(1.00 \mathrm{~atm}\). Given that \(C_{p, \mathrm{~m}}=R, \mu=0.21\) \(\mathrm{K} \mathrm{atm}^{-1}, B=-0.525(\mathrm{~K} / T) \mathrm{atm}^{-1}\), and that these are constant over the temperature range involved, calculate \(\Delta T\) and \(\Delta S\) for the gas.

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