(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, \(\left.-1.35 \times 10^{7} \mathrm{~Pa} / \mathrm{K} ?\right)\)

Step by step solution

01

Part (a) - Start with the Maxwell Relations

From the Maxwell relations for thermodynamic potentials, for a single-phase system, we have the relation: \[ \frac{\text{d}P}{\text{d}T} \bigg|_S = \frac{\text{d}(\frac{\text{d}F}{\text{d}V})_T}{\text{d}V} \bigg|_T \bigg|_S \bigg) \]. This tells us that we need to use the Helmholtz free energy where \(F = U - TS\).

02

Express Gibbs Free Energy in Terms of Partial Derivatives

Recall that Gibbs free energy, \(G\), is given by \(G = H - TS\). From this, we note that: \[ \text{d}G = V \text{d}P + S \text{d}T \]. For the entropy term we have: \[ \frac{\text{d}H}{\text{d}T}\bigg|_P = C_P \].

03

Utilize the Relation of Specific Heat Capacity at Constant Pressure (c_P)

Given that \(c_P = T(\frac{\text{d}S}{\text{d}T})_P\), we derive: \[ \frac{\text{d}S}{\text{d}T}\bigg|_P = \frac{c_P}{T} \].

04

Incorporate the Thermal Expansion Coefficient (Beta)

Consider the definition of thermal expansion coefficient \(\beta = \frac{1}{v} (\frac{\text{d}v}{\text{d}T})_P\). This is used in rearranging: \[ \frac{\text{d}v}{v} = \beta \text{d}T \].

05

Combine the Above Equations

Putting all the derived relationships together we ultimately achieve: \[ \bigg(\frac{\partial P}{\partial T}\bigg)_S = \frac{c_P}{T v \beta} \].

06

Part (b) - Substitute Given Values

Given parameters are: \(c_P = 2.01 \text{kJ/kg.K}\), \(v = 1.09 \times 10^{-3} \text{m}^3/\text{kg}\), \(\beta = 1.58 \times 10^{-4} \text{K}^{-1}\). Calculate: \[ \bigg(\frac{\partial P}{\partial T}\bigg)_S = \frac{2.01 \times 10^3}{(270.15 \times 1.09 \times 10^{-3} \times 1.58 \times 10^{-4})} = 4.31 \times 10^7 \text{Pa/K} \].

07

Part (c) - Calculate the Intersection Point

The equation of the adiabatic process has the same slope as \(\bigg(\frac{\partial P}{\partial T}\bigg)_S = 4.31 \times 10^{7} \text{Pa/K}\). Set this equal to the slope of the fusion curve: \[4.31 \times 10^{7} (T - 270.15) = P - 1.01 \times 10^{5}\]. Replacing \(T\) with known values at pressure points, solve for \(T\).

08

Find the Pressure at Melting Point

Use the temperature value calculated to plug back into the function of melting curve to find the pressure at the melting point.

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Key Concepts

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

Maxwell Relations

Maxwell Relations are a set of equations derived from the second derivatives of thermodynamic potentials. Using the potentials and the fact that mixed partial derivatives are equal, we derive four main Maxwell relations. These equations serve as bridges between different thermodynamic variables.
Each of these relations helps in translating one type of experimental measurement into another form. For instance, in our problem, we use thermodynamic potentials to relate derivatives involving entropy and volume into a more convenient form involving pressure and temperature.
Understanding and mastering Maxwell Relations is crucial because they simplify complex thermodynamic problems and offer a more intuitive grasp of the behavior of thermodynamic systems.

Helmholtz Free Energy

Helmholtz Free Energy (denoted as F) is one of the essential thermodynamic potentials used to describe the work obtainable from a thermodynamic system at constant temperature and volume. It is defined as:
\(F = U - TS\)
where U is internal energy, T is temperature, and S is entropy.
The differential form of Helmholtz Free Energy is given by:
\(\text{d}F = -S \text{d}T - P \text{d}V\).
This form is particularly useful because it allows for the derivation of one of the Maxwell Relations used in solving the original exercise. When studying changes involving Helmholtz Free Energy, it's essential to keep temperature and volume constant, making it helpful in many practical problems encountered in statistical mechanics and thermodynamics.

Gibbs Free Energy

Gibbs Free Energy (G) is another fundamental thermodynamic potential and is especially useful in processes occurring at constant pressure and temperature. It is given by:
\(G = H - TS\)
where H is enthalpy, T is temperature, and S is entropy.
The differential form of Gibbs Free Energy is:
\(\text{d}G = V \text{d}P + S \text{d}T\).
This equation indicates how Gibbs Free Energy changes with variations in pressure and temperature. Utilizing Gibbs Free Energy helps predict spontaneity and equilibrium conditions in chemical processes, determining whether a reaction will proceed or not. In our problem, understanding Gibbs Free Energy helps relate specific heat capacity term of entropy to derive necessary relationships.

Specific Heat Capacity

Specific Heat Capacity at constant pressure, denoted as \(c_P\), measures the amount of heat necessary to increase the temperature of a unit mass of a substance by one degree while keeping the pressure constant. For our problem, it converts the temperature change into entropy change as per:
\(\frac{\text{d}H}{\text{d}T}\bigg|_P = C_P\)
where \(H\) represents enthalpy.
Leveraging this specific heat, we can derive relations such as
\(\frac{\text{d}S}{\text{d}T}\bigg|_P = \frac{c_P}{T}\).
Specific heat capacities are crucial in interpreting how a material will respond to heating and are therefore important in engineering and thermodynamic calculations.

Thermal Expansion Coefficient

The Thermal Expansion Coefficient, denoted as \(\beta\), measures the fractional change in volume per degree change in temperature at constant pressure. It is defined as:
\( \beta = \frac{1}{V} \bigg(\frac{\text{d}V}{\text{d}T}\bigg)_P\).
In the exercise, \(\beta\) translates a small temperature change into a corresponding volume change, indicative of the material's ability to expand or contract. This concept fits snugly into our derivation thusly:
\(\frac{\text{d}v}{v} = \beta \text{d}T\).
Understanding the coefficient of thermal expansion is fundamental when dealing with materials subjected to temperature fluctuations, aiding in the design and analysis of various engineering systems.