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Chapter 3: Problem 21

Estimate the work necessary to compress the air in an air-compressor cylinder from a pressure of \(100 \mathrm{kPa}\) to \(2000 \mathrm{kPa}\). The initial volume is \(1000 \mathrm{~cm}^{3}\). An isothermal process is to be assumed.

Short Answer

Expert verified
The work necessary is approximately 36.02 J.

Step by step solution

01

Recall the formula for isothermal work

For an isothermal process, the work done on a gas is given by the formula \[ W = nRT \times \text{ln} \frac{P_2}{P_1} \] where \(P_1\) is the initial pressure, \(P_2\) is the final pressure, \(n\) is the number of moles of the gas, \(R\) is the universal gas constant, and \(T\) is the temperature.
02

Express number of moles using the ideal gas law

From the ideal gas law \( PV = nRT \), we rearrange to find the number of moles \(n\): \[ n = \frac{P_1V_1}{RT} \] where \(V_1\) is the initial volume. Let's first convert the volume from \( \text{cm}^3 \) to \( \text{m}^3 \). \[ V_1 = 1000 \text{ cm}^3 = 1000 \times 10^{-6} \text{ m}^3 = 0.001 \text{ m}^3 \]
03

Substitute values to find the number of moles \(n\)

With \(P_1 = 100 \text{ kPa} = 100 \times 10^3 \text{ Pa} \), substitute the values into the rearranged ideal gas law: \[ n = \frac{ (100 \times 10^3 \text{ Pa}) \times (0.001 \text{ m}^3) }{ (8.314 \text{ J/molK}) \times T } \] This simplifies to: \[ n = \frac{ 100 }{ 8.314 \times T } \]
04

Substitute \(n\) into the isothermal work formula

Next, we substitute the expression for \(n\) into the isothermal work formula: \[ W = \frac{100}{8.314} \times T \times \text{ln} \frac{2000 \text{ kPa}}{100 \text{ kPa}} \]
05

Simplify the expression

The temperature \(T\) cancels out, and we simplify the expression: \[ W = \frac{100}{8.314} \times \text{ln} 20 \]
06

Compute the natural logarithm and final work

Using \( \text{ln} 20 \approx 2.9957 \): \[ W = \frac{100}{8.314} \times 2.9957 \approx 36.02 \text{ J} \]

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Key Concepts

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

Isothermal Process
An isothermal process is one where the temperature of the system remains constant.
This implies that any heat added to the system is used entirely for doing work.
For our problem, this means that the air compressor maintains a constant temperature while compressing the air from 100 kPa to 2000 kPa.
This simplification helps because it allows us to use specific formulas to calculate the work done.
Ideal Gas Law
The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas.
It is expressed by the equation: \text{Ideal Gas Law:} \ PV = nRT
where:
  • \(P\) is the pressure of the gas
  • \(V\) is the volume of the gas
  • \(n\) is the number of moles
  • \(R\) is the universal gas constant (8.314 J/mol路K)
  • \(T\) is the temperature in Kelvin
This equation helps us understand the relationship between these variables and is crucial for finding the number of moles of gas when other quantities are known.
In this problem, we rearrange the ideal gas law to solve for the number of moles:
\(n = \frac{P_1V_1}{RT}\).
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a special kind of logarithm with the base \(e\), where \(e\) is approximately 2.718.
The natural logarithm is useful for solving problems in calculus and exponential growth/decay.
In our problem, it helps us calculate the work done during the isothermal compression.
Specifically, we use the term \(\ln\frac{P_2}{P_1}\) to find the log ratio of the final to initial pressures.
This ratio is essential in determining how much work is done on the system.
Pressure Conversion
Pressure is a measure of force applied over an area and can be represented in various units such as Pascals (Pa), Kilopascals (kPa), or Atmospheres (atm).
For this exercise, all given pressures are in kilopascals (kPa).
To use the ideal gas law and other formulas, it's critical to convert these pressures to Pascals (Pa).
For instance, converting 100 kPa to Pa is done using:
\(100 \text{ kPa} = 100\times10^3 \text{ Pa} \).
Converting to the same unit ensures consistency throughout the calculations.
Volume Conversion
Volume measures the space occupied by a gas and can be represented in units like cubic centimeters (\(cm^3\)), liters, or cubic meters (\(m^3\)).
In our problem, the initial volume of 1000 \(cm^3\) needs to be converted to cubic meters for use in calculations.
This conversion helps align units with the ideal gas law and other formulas.
The conversion process is straightforward:
\(1000 \text{ cm}^3 = 1000 \times 10^{-6} \text{ m}^3 = 0.001 \text{ m}^3 \).
Ensuring all volumes are in consistent units simplifies the math and reduces potential errors.

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Most popular questions from this chapter

A house has a \(0.5\)-cm-thick single-pane glass window \(2 \mathrm{~m}\) by \(1.5 \mathrm{~m}\). The inside temperature is \(20^{\circ} \mathrm{C}\) and the outside temperature is \(-20^{\circ} \mathrm{C}\). If there is an air layer on both the inside and the outside of the glass, each with an \(R\)-factor of \(0.1 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), determine the heat transfer rate through the window if \(k_{\text {glass }}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Derive an expression for the work required to stretch an unstretched length of wire a relatively small distance \(l\). The force is related to the amount of stretch \(x\) by \(F=E A x / L\), where \(L\) is the original length of the wire. \(A\) is the cross-sectional area, and \(E\) is a material constant, and modulus of elasticity.

An electronic gizmo with a surface area of \(75 \mathrm{~mm}^{2}\) generates \(3 \mathrm{~W}\) of heat. It is cooled by convection to air maintained at \(25^{\circ} \mathrm{C}\). If the surface temperature of the gizmo cannot exceed \(120^{\circ} \mathrm{C}\), estimate the heat transfer coefficient needed.

A linear spring requires \(20 \mathrm{~J}\) of work to compress it from an unstretched length of \(100 \mathrm{~mm}\) to a length of \(20 \mathrm{~mm}\). Find the spring constant.

Two kilograms of saturated steam at \(400 \mathrm{kPa}\) is contained in a piston-cylinder arrangement. The steam is heated at constant pressure to \(300^{\circ} \mathrm{C}\). Calculate the work done by the steam.
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