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The Tait equation for liquids is written for an isotherm as: $$V=V_{0}\left(1-\frac{A P}{B+P}\right)$$ where \(V\) is molar or specific volume, \(V_{0}\) is the hypothetical molar or specific volume at zero pressure, and \(\mathrm{A}\) and \(\mathrm{B}\) are positiveconstants. Find an expressionfor the isothermal compressibility consistent with this equation.

Short Answer

Expert verified
\( \kappa_T = \frac{V_{0} AB}{V (B+P)^2} \).

Step by step solution

01

Understanding Isothermal Compressibility

The isothermal compressibility, denoted as \( \kappa_T \), is a measure of the relative volume change of a liquid or solid as a response to a pressure change at constant temperature. Mathematically, it is defined as: \[ \kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T \] This definition will guide us in finding the derivative of volume \( V \) with respect to pressure \( P \).
02

Differentiate the Tait Equation with Respect to Pressure

Using the given Tait equation: \[ V = V_{0}\left(1-\frac{A P}{B+P}\right) \] Differentiate \( V \) with respect to \( P \): \[ \frac{\partial V}{\partial P} = V_{0}\left( -\frac{A (B+P) - A P}{(B+P)^2} \right) \] This simplifies to: \[ \frac{\partial V}{\partial P} = V_{0} \left(-\frac{AB}{(B+P)^2}\right) \]
03

Substitute the Derivative into the Compressibility Formula

Now that we have \( \frac{\partial V}{\partial P} \), substitute it into the isothermal compressibility formula: \[ \kappa_T = -\frac{1}{V}\left( \frac{\partial V}{\partial P} \right)_T = -\frac{1}{V} V_{0} \left(-\frac{AB}{(B+P)^2}\right) \] Simplifying this expression gives us: \[ \kappa_T = \frac{V_{0} AB}{V (B+P)^2} \]
04

Express \( V \) from the Tait Equation

We know from the original equation that: \[ V = V_{0}\left( 1-\frac{A P}{B+P} \right) \] Using this, express \( V \) in terms of known constants and \( P \): \[ V = V_{0} - \frac{V_{0} AP}{B+P} \]
05

Conclude with the Isothermal Compressibility Expression

Using the expression for \( V \) from the Tait equation in \( \kappa_T \) gives us the final form: \[ \kappa_T = \frac{V_{0} AB}{(V_{0} - \frac{V_{0} AP}{B+P}) (B+P)^2} \] This is the expression for the isothermal compressibility consistent with the Tait equation.

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Key Concepts

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

Tait Equation
The Tait equation is used to describe the behavior of liquids under different pressures while maintaining the same temperature (isothermally). It provides a relationship between the molar volume or specific volume of a liquid and the pressure it is under. The equation is given as:\[ V = V_{0}\left(1-\frac{A P}{B+P}\right) \]where:
  • \( V \) is the molar or specific volume of the liquid.
  • \( V_{0} \) is the molar or specific volume at zero pressure.
  • \( A \) and \( B \) are constants unique to each liquid.
  • \( P \) is the pressure applied to the liquid.
The equation helps to predict how a liquid's volume will change as external pressure changes. This is fundamental in understanding fluid mechanics and engineering applications involving fluids, such as hydraulic systems and pressure vessels.
Molar Volume
Molar volume refers to the volume occupied by one mole of a substance, be it a solid, liquid, or gas. It is an important concept in chemistry and thermodynamics, as it provides insight into the arrangement of molecules in a substance.In the context of the Tait equation, molar volume \( V \) is used to measure how a liquid's volume changes under pressure. The Tait equation expresses the molar volume as a function of pressure and other constants. When dealing with liquids, particularly those in different states or under varying environmental conditions, molar volume calculations can help predict behavior in reaction processes or phase changes. A liquid's responsiveness to pressure, and its subsequent volume change, makes this concept crucial for material and chemical engineers.
Pressure Differentiation
Pressure differentiation involves calculating how a liquid's properties, particularly its volume, change with varying pressure. In this exercise, the goal was to derive the expression for the isothermal compressibility using pressure differentiation of the Tait equation.Differentiating involves finding the gradient at which volume \( V \) changes with respect to pressure \( P \). This differential is given as:\[ \frac{\partial V}{\partial P} = V_{0}\left(-\frac{AB}{(B+P)^2}\right) \]This expression indicates that volume change is dependent on constants \( A \) and \( B \), as well as the initial molar volume \( V_0 \) and current pressure \( P \). Understanding how these factors interact aids in predicting and controlling liquid behavior under pressure, which is vital in many industrial processes.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships and conversions between heat and other forms of energy. It is a foundational principle in understanding how various processes occur in nature, including those involving fluids and gases.In the context of this exercise, thermodynamics comes into play with the isothermal process, which refers to changes occurring at a constant temperature. Isothermal compressibility, defined as:\[ \kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T \]provides a measure of how much the volume of a liquid changes under pressure at a constant temperature. This property is crucial for accurately describing and predicting a system's response to external pressure changes.Real-life applications of thermodynamics in fluids include natural phenomena like ocean currents and human-engineered solutions, including power plants, refrigeration, and heating systems.

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Most popular questions from this chapter

A \(0.35-\mathrm{m}^{3}\) vessel holds ethane vapor at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and \(2200 \mathrm{kPa}\). If it is heated to \(493.15 \mathrm{K}\left(220^{\circ} \mathrm{C}\right),\) what pressure is developed?

Calculate the reversible work done in compressing \(0.0283 \mathrm{m}^{3}\) of mercury at a constant temperature of \(273.15 \mathrm{K}\left(\mathrm{O}^{\circ} \mathrm{C}\right)\) from 1 atm to 3000 atm. The isothermal compressibility of mercury at \(273.15 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is $$\kappa=3.9 \times 10^{-6}-0.1 \times 10^{-9} P$$ where \(P\) is in atm and \(\mathrm{K}\) is in \(\mathrm{atm}^{-1}\).

Express the volume expansivity and the isothermal compressibility as functions of density \(\rho\) and its partialderivatives.For water at \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\) and 1 bar, \(\mathrm{K}=44.18 \times 10^{-6}\) bar \(^{-1}\). To what pressure must water be compressed at \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\) to change its density by \(1 \%\) ? Assume that \(\kappa\) is independent of \(\mathbf{P}\).

A process consists of two steps: (1) One \(\mathrm{kmol}\) of air at \(\mathrm{T}=800 \mathrm{K}\) and \(P=4\) bar is cooled at constant volume to \(\mathrm{T}=350 \mathrm{K}\). ( 2 ) The air is then heated at constant pressure until its temperature reaches \(800 \mathrm{K}\). If this two-step process is replaced by a single isothermal expansion of the air from \(800 \mathrm{K}\) and 4 bar to some final pressure \(\mathrm{P}\), what is the value of \(P\) that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with \(C_{P}=(7 / 2) R\) and \(C_{V}=(5 / 2) R\).

\(\mathbf{A}\) tank of \(0.1-\mathrm{m}^{3}\) volume contains air at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and \(101.33 \mathrm{kPa}\). The tank is connected to a compressed-air line which supplies air at the constant conditions of \(318.15 \mathrm{K}\left(45^{\circ} \mathrm{C}\right)\) and \(1500 \mathrm{kPa} .\) A valve in the line is cracked so that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperaturein the tank remains at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\), how much heat is lost from the tank? Assume air to be an ideal gas for which \(C_{P}=(7 / 2) R\) and \(C_{V}=(5 / 2) R\).

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