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Chapter 19: Problem 8

An ideal gas undergoes a process during which the pressure is kept directly proportional to the volume, so that \(p=\alpha V\) where \(\alpha\) is a positive constant. If the volume changes from \(V_{1}\) to \(V_{2}\), how much work is done by the gas? Express your answer in terms of \(V_{1}, V_{2},\) and \(\alpha\)

Short Answer

Expert verified
The work done by the gas is \(W = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)

Step by step solution

01

Writing the Work Formula

The work done \(W\) on/by the gas can be given by the integral of pressure \(p\) with respect to volume \(V\), from the initial volume \(V_{1}\) to the final volume \(V_{2}\). The formula is \(W = \int_{V_{1}}^{V_{2}} pdV\).
02

Substituting the Pressure-Volume Relationship

Since the pressure is directly proportional to the volume, \(p = \alpha V\). By substituting this into the work formula, the equation becomes: \(W = \int_{V_{1}}^{V_{2}} \alpha V dV\)
03

Calculating the Integral

The integral can be calculated by the power rule: \(\int x^n dx = \frac{1}{n+1}x^{n+1}\). Therefore, \( \int_{V_{1}}^{V_{2}} \alpha V dV = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)
04

Final Answer

The work done by the gas is given by: \(W = \frac{1}{2} \alpha (V_{2}^2 - V_{1}^2)\)

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Key Concepts

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

Work Done by Gas
When we talk about the work done by a gas, we're looking at how much energy the gas has expended while changing its volume. This is a fundamental concept in thermodynamics where gases do work on their surroundings by expanding. For an ideal gas undergoing a transformation, the work done can be calculated by the integral of pressure with respect to volume change. In simpler terms, if you imagine a piston filled with gas, as the gas expands and pushes the piston outward, it's doing work against the external pressure.

In our specific scenario, the gas鈥� pressure increases proportionally to its volume, which makes for an interesting case as work calculation involves integrating a function where pressure is a variable that depends on volume. Remember that the work done by a gas can be both positive and negative鈥攑ositive when the gas expands (doing work on the surroundings) and negative when it's compressed (work is done on the gas).
Pressure-Volume Relationship
The pressure-volume relationship, often depicted on a graph as a curve, is crucial in understanding how gases behave under different conditions. For an ideal gas, the product of pressure and volume is a constant when temperature remains unchanged (Boyle鈥檚 Law). However, in our exercise, we鈥檙e looking at a direct proportionality; as the volume, denoted by V, increases, so does the pressure, represented by p. This relationship can be mathematically expressed as p = \(\alpha V\), where \(\alpha\) is a constant.

This linear relationship between pressure and volume means that as the gas expands in a container, the pressure it exerts on the container walls increases at the same rate. This sets the stage to calculate work done by using integral calculus, as the work is essentially the area under the curve represented by this linear relationship in a pressure-volume graph.
Integral Calculus in Physics
The application of integral calculus in physics is a powerful tool for solving a variety of problems, particularly those involving changes over an interval. Integrals allow you to calculate the total effect of something that changes across a range, like how the area under a pressure-volume curve represents the work done by a gas during expansion or compression.

In this exercise, integral calculus is used to determine the work done by the gas when the pressure and volume are directly proportional. A definite integral from V1 to V2 is needed to find the exact amount of work done over the change in volume. The integral of the function p = \(\alpha V\) from the initial to the final volume is established and, with the application of the power rule, we can compute the answer. By integrating, we sum up infinitely small amounts of work done by the gas during each infinitely small increase in volume, giving us the total work done over the entire process.

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Most popular questions from this chapter

Helium gas undergoes an adiabatic process in which the Kelvin temperature doubles. By what factor does the pressure change?

A cylinder with a piston contains \(0.150 \mathrm{~mol}\) of nitrogen at \(1.80 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K}\). The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(p V\) -diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

\(\cdot\) A cylinder contains \(0.0100 \mathrm{~mol}\) of helium at \(T=27.0^{\circ} \mathrm{C}\) (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C} ?\) Draw a \(p V\) -diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

In a hospital, pure oxygen may be delivered at 50 psi (gauge pressure) and then mixed with \(\mathrm{N}_{2} \mathrm{O}\). What volume of oxygen at \(20^{\circ} \mathrm{C}\) and 50 psi (gauge pressure) should be mixed with \(1.7 \mathrm{~kg}\) of \(\mathrm{N}_{2} \mathrm{O}\) to get a \(50 \% / 50 \%\) mixture by volume at \(20^{\circ} \mathrm{C} ?\) (a) \(0.21 \mathrm{~m}^{3} ;\) (b) \(0.27 \mathrm{~m}^{3}\); (c) \(1.9 \mathrm{~m}^{3} ;\) (d) \(100 \mathrm{~m}^{3}\).

A steel cargo drum has a height of \(880 \mathrm{~mm}\) and a diameter of \(610 \mathrm{~mm}\). With its top removed it has a mass of \(17.3 \mathrm{~kg}\). The drum is turned upside down at the surface of the North Atlantic and is pulled downward into the ocean by a robotic submarine. On this day the surface temperature is \(23.0^{\circ} \mathrm{C}\) and the surface air pressure is \(p_{0}=101 \mathrm{kPa}\). The water temperature decreases linearly with depth to \(3.0^{\circ} \mathrm{C}\) at \(1000 \mathrm{~m}\) below the surface. As the drum moves downward in the ocean, the air inside the drum is compressed, reducing the upward buoyant force. (a) At what depth \(y_{\text {neutral }}\) is the barrel neutrally buoyant? (Hint: The pressure in the drum is equal to the sea pressure, which at depth \(y\) is \(p_{0}+\rho g y\) where \(\rho=1025 \mathrm{~kg} / \mathrm{m}^{3}\) is the density of seawater. The temperature at depth can be determined using the information above. Together with the ideal- gas law, you can derive a formula for the volume of air at depth \(y,\) and therefore a formula for the upward buoyant force as a function of depth.) (b) What is the volume of the air in the drum at depth \(y_{\text {neutral }} ?\)
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