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Chapter 11: Problem 100

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T\) , where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\) .

Short Answer

Expert verified
(a) Bulk modulus is equal to pressure (\( B = P \)) for isothermal compression. (b) Bulk modulus is \( \gamma P \) for adiabatic compression.

Step by step solution

01

Understanding Bulk Modulus for Isothermal Process

The bulk modulus \( B \) is defined as \( B = -V \frac{\Delta P}{\Delta V/V} \), which is the negative ratio of the volume and the fractional change in volume under pressure change. For isothermal compression, \( T \) is constant and thus, \( PV = nRT \) implies \( P \) is inversely proportional to \( V \). Differentiating \( PV = nRT \) w.r.t volume \( V \), keeping \( T \) constant, we get \( P dV + V dP = 0 \) or \( dP = -\frac{P}{V} dV \). Substituting in \( B = -V \frac{dP}{dV/V} \), we simplify: \( B = P \). Therefore, the bulk modulus equals the pressure \( P \).
02

Understanding Bulk Modulus for Adiabatic Process

In an adiabatic process, no heat is transferred (\( dQ = 0 \)), and pressure \( P \) and volume \( V \) are related by \( PV^\gamma = \text{constant} \). Differentiating this equation with respect to \( V \), we find \( \frac{dP}{dV} = -\gamma \frac{P}{V} \). Substituting this into \( B = -V \frac{dP}{dV/V} \), gives \( B = \gamma P \). Therefore, when gas compression is adiabatic, the bulk modulus becomes \( \gamma \) times the pressure \( P \).

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Key Concepts

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

Bulk Modulus
The bulk modulus is a measure of a material's resistance to uniform compression. It tells us how hard it is to compress a gas. In terms of an ideal gas, it connects pressure and volume changes.
- Mathematically, the bulk modulus \( B \) is defined as \( B = -V \frac{\Delta P}{\Delta V/V} \). Here, \( \Delta P \) is the change in pressure, and \( \Delta V \) is the change in volume. The negative sign indicates that an increase in pressure usually decreases volume.
- For an isothermal process (constant temperature), the ideal gas law \( PV = nRT \) implies that temperature \( T \) is constant. When differentiated with respect to \( V \), it gives us \( P \) is inversely proportional to \( V \). This means an increase in pressure leads to a decrease in volume, or vice versa.
- Upon simplifying, the bulk modulus in an isothermal process equals the current pressure \( P \) at any point. Knowing that \( B = P \) makes it easier to predict how the gas will behave under specific pressure conditions.
Isothermal Process
An isothermal process is one where the temperature remains constant throughout the process.
- For an ideal gas undergoing an isothermal process, the equation \( PV = nRT \) comes into play. Since \( T \) is constant, \( P \) and \( V \) adjust to keep the product \( PV \) unchanged.
- In practical terms, when you compress an ideal gas isothermally, you must slowly release the heat energy to keep temperature from rising. This keeps the temperature steady even as volume changes.
- From a mathematical perspective, for isothermal compression, the differentials \( P dV + V dP = 0 \) simplify to show \( dP = -\frac{P}{V} dV \). Plugging this into the bulk modulus formula results in \( B = P \), showing the balance between pressure and volume under constant temperature conditions.
Adiabatic Process
An adiabatic process occurs when no heat is exchanged between the system and its surroundings.
- For an ideal gas in an adiabatic process, the relation \( PV^\gamma = \text{constant} \) is used, where \( \gamma \) is the adiabatic index. Different gases have different values of \( \gamma \). This equation describes how pressure \( P \) and volume \( V \) vary when the gas is compressed or expanded without heat flow.
- When you differentiate this relationship with respect to volume, it yields \( \frac{dP}{dV} = -\gamma \frac{P}{V} \). Substituting this into the bulk modulus expression \( B = -V \frac{dP}{dV/V} \) simplifies to \( B = \gamma P \).
- This means that in an adiabatic compression, the bulk modulus is \( \gamma \) times the pressure, which reflects the increased resistance to compression because no heat is dissipated during the process. This is different from isothermal processes, where temperature remains constant and bulk modulus equals the pressure.

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Most popular questions from this chapter

A \(350-N,\) uniform, \(1.50-m\) bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 \(\mathrm{N}\) without breaking, and cable \(B\) can support up to 400.0 \(\mathrm{N}\) . You want to place a small weight on this bart up is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

A \(1.05-\mathrm{m}\) -long rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. 11.62\() .\) The crosssectional arca of \(A\) is 2.00 \(\mathrm{mm}^{2}\) und that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) At what point along the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B,\) and \((b)\) equal strains in \(A\) and \(B ?\)

A nonuniform beam 4.50 \(\mathrm{m}\) long and weighing 1.00 \(\mathrm{kN}\) makes an angle of \(25.0^{\circ}\) below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 \(\mathrm{m}\) farther down the beam and perpendicular to it (Fig. 11.31\()\) . The center of gravity of the beam is 2.00 \(\mathrm{m}\) down the beam from the pivot. Lighting equipment exerts a \(5.00-\mathrm{kN}\) downward force on the lower left endof the beam. Find the tension \(T\) in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

Forearm. In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 \(\mathrm{cm}\) from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 \(\mathrm{N}\) and that their center of gravity is 15.0 \(\mathrm{cm}\) from the elbow (not quite halfway to the hand). The forearm is held horizontally at a a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the foreann. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds a \(80.0-\mathrm{N}\) weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 \(\mathrm{cm}\) from the elbow. Construct a freebody diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part ( \((b),\) find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the 80.0 -N weight, the person raises his forearm. until it is at an angle of \(53.0^{\circ}\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearmis in this position? Has the force increased or decreased fromits value in part (b)? Explain why this is so, and test your answer by actually doing this with your own arm.
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