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Chapter 15: Problem 7

The following terms are used in thermodynamics: Isothermal compressibility \(\kappa=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)\) Thermal expansion coefficient $$ \beta=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right) $$ Show that for the ideal gas equation, \(P V=R T\), we have \(\kappa=\frac{1}{P}\) and \(\beta=\frac{1}{T}\)

Short Answer

Expert verified
For the ideal gas equation, we have shown that \(\kappa = \frac{1}{P}\) and \(\beta = \frac{1}{T}\).

Step by step solution

01

Write Down the Ideal Gas Equation

The ideal gas equation is given as: \[ PV = RT \] where P is pressure, V is volume, R is the ideal gas constant, and T is temperature.
02

Derive Isothermal Compressibility (\(\kappa\))

First, we need to find \[ \frac{\partial V}{\partial P}\] while keeping temperature (T) constant (isothermal process).Start by rearranging the ideal gas equation:\[ V = \frac{RT}{P} \] Taking the partial derivative of V with respect to P:\[ \frac{\partial V}{\partial P} = \frac{\partial}{\partial P}\left(\frac{RT}{P}\right) = -\frac{RT}{P^{2}} \] Next, apply it in the definition of \(\kappa\): \[\kappa = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right) = -\frac{1}{\frac{RT}{P}}\left( -\frac{RT}{P^{2}} \right) = \frac{1}{P} \]
03

Derive Thermal Expansion Coefficient (\(\beta\))

Next, we find \[ \frac{\partial V}{\partial T}\] while keeping pressure (P) constant.Start with the same rearranged ideal gas equation:\[ V = \frac{RT}{P} \] Taking the partial derivative of V with respect to T:\[ \frac{\partial V}{\partial T} = \frac{\partial}{\partial T}\left(\frac{RT}{P}\right) = \frac{R}{P} \] Next, apply it in the definition of \(\beta\): \[ \beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right) = \frac{1}{\frac{RT}{P}} \left( \frac{R}{P} \right) = \frac{1}{T} \]
04

Conclusion

From the above derivations, we have shown that: \[ \kappa = \frac{1}{P} \] and \[ \beta = \frac{1}{T} \] for an ideal gas equation.

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Key Concepts

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

Isothermal Compressibility (魏)
Isothermal compressibility is a measure of the change in volume of a substance when subjected to pressure, while keeping the temperature constant. It is represented as:
\[ \kappa = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T \]
Here, \( V \) is volume, \( P \) is pressure, and the partial derivative \( \frac{\partial V}{\partial P} \) shows how much the volume changes with respect to a small change in pressure.
For an ideal gas, we use the ideal gas law \[ PV = RT \] where \( R \) is the ideal gas constant and \( T \) is temperature. By rearranging the ideal gas law to \[ V = \frac{RT}{P} \] and taking the partial derivative of \( V \) with respect to \( P \), we get: \[ \frac{\partial V}{\partial P} = -\frac{RT}{P^2} \]
Substituting this back into the definition of isothermal compressibility, we have:
\[ \kappa = -\frac{1}{\frac{RT}{P}} \left( -\frac{RT}{P^2} \right) = \frac{1}{P} \]
This shows that for an ideal gas, isothermal compressibility is inversely proportional to pressure.
Thermal Expansion Coefficient (尾)
The thermal expansion coefficient indicates how the volume of a substance changes with temperature at constant pressure. It is defined as:
\[ \beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P \]
For an ideal gas, using the ideal gas law \[ PV = RT \], we rearrange to get \[ V = \frac{RT}{P} \]. Taking the partial derivative of \( V \) with respect to \( T \), gives: \[ \frac{\partial V}{\partial T} = \frac{R}{P} \]
Applying this derivative to the definition of the thermal expansion coefficient, we get:
\[ \beta = \frac{1}{\frac{RT}{P}} \left( \frac{R}{P} \right) = \frac{1}{T} \] This result shows that for an ideal gas, the thermal expansion coefficient is inversely proportional to the temperature.
Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that describes the behavior of an ideal gas. The equation is: \[ PV = RT \]
Here,
  • \( P \) is pressure
  • \( V \) is volume
  • \( R \) is the ideal gas constant, approximately \( 8.314 \) J/(mol路K)
  • \( T \) is temperature

This equation states that the product of the pressure and volume of an ideal gas is directly proportional to its temperature. The ideal gas law is crucial for understanding the properties of gases and forms the basis for many other equations in thermodynamics.
By manipulating this equation, we derived expressions for other properties like isothermal compressibility (\

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Most popular questions from this chapter

[fluid mechanics] The rate of discharge, \(Q\), of a liquid through an orifice (opening) of area \(A\) is given by $$ Q=0.7 A \sqrt{2 g h} $$ where \(h\) is the depth of the centre of the orifice and \(g\) is the acceleration due to gravity. Given that \(A=\left(2 \times 10^{-3}\right) \mathrm{m}^{2}\), \(h=1.75 \mathrm{~m}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\), find the approximate change in \(Q\) if \(A\) is increased by \(\left(0.2 \times 10^{-3}\right) \mathrm{m}^{2}\) and \(h\) is decreased by \(0.25 \mathrm{~m}\). (The SI units of \(Q\) are \(\mathrm{m}^{3} / \mathrm{s}\).)

Find where the stationary points of the following functions occur: a \(f(x, y)=x y+\frac{1}{2} y^{2}-7 y-4 x\) b \(f(x, y)=x^{2}+y^{2}+4 x y-5 x-4 y\) c \(f(x, y)=x^{3}+y^{2}+x y+22 y\)

(structures] Given the stress function, \(\Omega(x, y)\), as (A and B are constants) $$ \Omega(x, y)=A x^{2} y^{4}+B x^{4} y^{2} $$ find the stresses, \(\sigma_{x}\) and \(\sigma_{y}\), where $$ \sigma_{x}=\frac{\partial^{2} \Omega}{\partial y^{2}} \text { and } \sigma_{y}=\frac{\partial^{2} \Omega}{\partial x^{2}} $$

X [aerodynamics] The lift, \(L\), of a body in a fluid of density \(\rho\) is given by $$ L=\frac{C \rho V^{2} A}{2} $$ where \(C\) is the lift coefficient, \(V\) is the free stream velocity and \(A\) is the area. If \(C\) can be measured to within \(1 \%, \rho\) to within \(0.5 \%, V\) to within \(0.6 \%\) and \(A\) to within \(0.1 \%\), find the largest percentage error in the value of \(L\).

equation is given by $$ T=\frac{P V}{R} $$ where \(T\) is the temperature, \(V\) is the volume, \(P\) is the pressure and \(R\) is the gas constant. Show that $$ \mathrm{d} T=\frac{V \mathrm{~d} P+P \mathrm{~d} V}{R} $$
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