what happens to the internal energy of a system when work is done on it

The Offset Law of Thermodynamics

fifteen Work, Heat, and Internal Energy

Learning Objectives

By the end of this section, you will exist able to:

Describe the work done by a organisation, rut transfer between objects, and internal free energy change of a system
Calculate the work, heat transfer, and internal energy modify in a simple process

We discussed the concepts of work and energy earlier in mechanics. Examples and related problems of heat transfer betwixt dissimilar objects have also been discussed in the preceding capacity. Here, we desire to aggrandize these concepts to a thermodynamic system and its surroundings. Specifically, we elaborated on the concepts of heat and heat transfer in the previous two chapters. Here, we want to understand how work is done by or to a thermodynamic organisation; how heat is transferred between a organisation and its environs; and how the total energy of the system changes under the influence of the work washed and oestrus transfer.

Work Washed past a System

A force created from whatsoever source can practice work by moving an object through a displacement. Then how does a thermodynamic system practise piece of work? (Effigy) shows a gas confined to a cylinder that has a movable piston at i end. If the gas expands against the piston, it exerts a force through a altitude and does work on the piston. If the piston compresses the gas as it is moved in, piece of work is also done—in this case, on the gas. The work associated with such book changes can be determined every bit follows: Permit the gas force per unit area on the piston confront be p. Then the forcefulness on the piston due to the gas is pA, where A is the expanse of the face. When the piston is pushed outward an minute distance dx, the magnitude of the work washed past the gas is

$dW=F\phantom{\rule{0.2em}{0ex}}dx=pA\phantom{\rule{0.2em}{0ex}}dx.$

Since the change in volume of the gas is $dV=A\phantom{\rule{0.2em}{0ex}}dx,$ this becomes

$dW=pdV.$

For a finite change in book from ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2},$ we can integrate this equation from ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2}$ to find the cyberspace work:

$W={\int }_{{V}_{1}}^{{V}_{2}}pdV.$

The work washed by a confined gas in moving a piston a distance dx is given by $dW=Fdx=pdV.$

The figure is an illustration of a piston with gas inside. The piston is shown in two positions, separated by a distance d x. A force F equal to p A is shown pushing the piston outward.

This integral is only meaningful for a quasi-static process, which means a process that takes place in infinitesimally small steps, keeping the arrangement at thermal equilibrium. (Nosotros examine this idea in more detail later in this affiliate.) Only then does a well-defined mathematical relationship (the equation of state) be between the pressure and volume. This relationship can exist plotted on a pV diagram of pressure versus volume, where the curve is the alter of state. We tin can gauge such a process as 1 that occurs slowly, through a series of equilibrium states. The integral is interpreted graphically as the area under the pV curve (the shaded area of (Figure)). Work washed past the gas is positive for expansion and negative for compression.

When a gas expands slowly from ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2},$ the work done by the system is represented by the shaded expanse under the pV curve.

The figure shows a graph of p on the vertical axis as a function of V on the horizontal axis. No scale or units are given for either axis. Two points are labeled: p 1, V 1 and p 2, V 2, with V 2 larger than V 1 and p 2 smaller than p 1. A curve connects the two points and the area under the curve is shaded. The curve is concave up.

Consider the ii processes involving an platonic gas that are represented by paths AC and ABC in (Effigy). The first process is an isothermal expansion, with the volume of the gas changing its volume from ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2}$ . This isothermal process is represented by the curve between points A and C. The gas is kept at a constant temperature T by keeping it in thermal equilibrium with a heat reservoir at that temperature. From (Figure) and the platonic gas constabulary,

$W={\int }_{{V}_{1}}^{{V}_{2}}pdV={\int }_{{V}_{1}}^{{V}_{2}}\left(\frac{nRT}{V}\right)\phantom{\rule{0.2em}{0ex}}dV.$

The paths ABC, AC, and ADC correspond three different quasi-static transitions between the equilibrium states A and C.

The expansion is isothermal, so T remains abiding over the entire process. Since due north and R are as well constant, the simply variable in the integrand is V, so the work washed past an ideal gas in an isothermal procedure is

$W=nRT{\int }_{{V}_{1}}^{{V}_{2}}\frac{dV}{V}=nRT\text{ln}\frac{{V}_{2}}{{V}_{1}}.$

Detect that if ${V}_{2}>{V}_{1}$ (expansion), W is positive, equally expected.

The straight lines from A to B and so from B to C represent a different procedure. Hither, a gas at a pressure ${p}_{1}$ first expands isobarically (constant pressure level) and quasi-statically from ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2}$ , afterwards which it cools quasi-statically at the constant volume ${V}_{2}$ until its pressure drops to ${p}_{2}$ . From A to B, the pressure is constant at p, so the work over this part of the path is

$W={\int }_{{V}_{1}}^{{V}_{2}}pdV={p}_{1}{\int }_{{V}_{1}}^{{V}_{2}}dV={p}_{1}\left({V}_{2}-{V}_{1}\right).$

From B to C, there is no change in volume and therefore no work is washed. The cyberspace work over the path ABC is then

$W={p}_{1}\left({V}_{2}-{V}_{1}\right)+0={p}_{1}\left({V}_{2}-{V}_{1}\right).$

A comparison of the expressions for the work done by the gas in the two processes of (Figure) shows that they are quite different. This illustrates a very important belongings of thermodynamic work: It is path dependent. We cannot determine the piece of work done past a organisation every bit it goes from one equilibrium state to some other unless we know its thermodynamic path. Different values of the piece of work are associated with unlike paths.

Isothermal Expansion of a van der Waals Gas Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gas molecules have a definite volume (see The Kinetic Theory of Gases). One mole of a van der Waals gas has an equation of state

$\left(p+\frac{a}{{V}^{2}}\right)\left(V-b\right)=RT,$

where a and b are ii parameters for a specific gas. Suppose the gas expands isothermally and quasi-statically from volume ${V}_{1}$ to volume ${V}_{2}.$ How much work is washed by the gas during the expansion?

Strategy Considering the equation of state is given, we can use (Figure) to limited the pressure in terms of V and T. Furthermore, temperature T is a constant under the isothermal condition, and so V becomes the only changing variable nether the integral.

Solution To evaluate this integral, we must express p as a function of V. From the given equation of state, the gas pressure is

$p=\frac{RT}{V-b}-\frac{a}{{V}^{2}}.$

Considering T is constant nether the isothermal condition, the piece of work done by ane mol of a van der Waals gas in expanding from a volume ${V}_{1}$ to a volume ${V}_{2}$ is thus

$\begin{array}{cc}\hfill W& =\underset{{V}_{1}}{\overset{{V}_{2}}{\int }}\left(\frac{RT}{V-b}-\frac{a}{{V}^{2}}\right)={|RT\text{ln}\left(V-b\right)+\frac{a}{V}|}_{{V}_{1}}^{{V}_{2}}\hfill \\ & =RT\text{ln}\left(\frac{{V}_{2}-b}{{V}_{1}-b}\right)+a\left(\frac{1}{{V}_{2}}-\frac{1}{{V}_{1}}\right).\hfill \end{array}$

Significance By taking into business relationship the volume of molecules, the expression for work is much more complex. If, nevertheless, we set $a=0$ and $b=0,$ we encounter that the expression for work matches exactly the piece of work done by an isothermal process for one mole of an ideal gas.

Check Your Understanding How much work is done by the gas, as given in (Figure), when it expands quasi-statically along the path ADC?

${p}_{2}\left({V}_{2}-{V}_{1}\right)$

Internal Free energy

The internal energy ${E}_{\text{int}}$ of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the organization. If the kinetic and potential energies of molecule i are ${K}_{i}$ and ${U}_{i},$ respectively, then the internal energy of the organisation is the average of the total mechanical energy of all the entities:

${E}_{\text{int}}=\sum _{i}\left({\stackrel{\text{−}}{K}}_{i}+{\stackrel{\text{−}}{U}}_{i}\right),$

where the summation is over all the molecules of the system, and the bars over One thousand and U bespeak average values. The kinetic energy ${K}_{i}$ of an individual molecule includes contributions due to its rotation and vibration, as well equally its translational free energy ${m}_{i}{v}_{i}^{2}\text{/}2,$ where ${v}_{i}$ is the molecule's speed measured relative to the center of mass of the organisation. The potential energy ${U}_{i}$ is associated merely with the interactions between molecule i and the other molecules of the system. In fact, neither the arrangement'due south location nor its motility is of any consequence as far as the internal energy is concerned. The internal energy of the organisation is not affected past moving it from the basement to the roof of a 100-story building or past placing it on a moving railroad train.

In an ideal monatomic gas, each molecule is a unmarried atom. Consequently, there is no rotational or vibrational kinetic free energy and ${K}_{i}={m}_{i}{v}_{i}^{2}\text{/}2$ . Furthermore, there are no interatomic interactions (collisions withal), and then ${U}_{i}=\text{constant}$ , which we set to zero. The internal energy is therefore due to translational kinetic energy just and

${E}_{\text{int}}=\sum _{i}{\stackrel{\text{−}}{K}}_{i}=\sum _{i}\frac{1}{2}{m}_{i}\overline{{v}_{i}^{2}}.$

From the discussion in the preceding affiliate, we know that the average kinetic energy of a molecule in an platonic monatomic gas is

$\frac{1}{2}{m}_{i}\stackrel{\text{−}}{{v}_{i}^{2}}=\frac{3}{2}{k}_{\text{B}}T,$

where T is the Kelvin temperature of the gas. Consequently, the boilerplate mechanical free energy per molecule of an ideal monatomic gas is also $3{k}_{\text{B}}T\text{/}2,$ that is,

$\overline{{K}_{i}+{U}_{i}}=\stackrel{\text{−}}{{K}_{i}}=\frac{3}{2}{k}_{\text{B}}T.$

The internal free energy is simply the number of molecules multiplied past the average mechanical energy per molecule. Thus for n moles of an ideal monatomic gas,

${E}_{\text{int}}=n{N}_{\text{A}}\left(\frac{3}{2}{k}_{\text{B}}T\right)=\frac{3}{2}nRT.$

Notice that the internal energy of a given quantity of an platonic monatomic gas depends on but the temperature and is completely independent of the pressure and volume of the gas. For other systems, the internal energy cannot be expressed so simply. All the same, an increase in internal energy can often be associated with an increment in temperature.

Nosotros know from the zeroth law of thermodynamics that when two systems are placed in thermal contact, they eventually reach thermal equilibrium, at which point they are at the same temperature. Every bit an case, suppose nosotros mix ii monatomic ideal gases. Now, the free energy per molecule of an ideal monatomic gas is proportional to its temperature. Thus, when the two gases are mixed, the molecules of the hotter gas must lose energy and the molecules of the colder gas must gain energy. This continues until thermal equilibrium is reached, at which point, the temperature, and therefore the average translational kinetic energy per molecule, is the same for both gases. The arroyo to equilibrium for real systems is somewhat more complicated than for an ideal monatomic gas. Nevertheless, nosotros can still say that energy is exchanged between the systems until their temperatures are the same.

Summary

Positive (negative) work is washed by a thermodynamic system when it expands (contracts) under an external pressure level.
Estrus is the energy transferred between ii objects (or 2 parts of a system) because of a temperature difference.
Internal energy of a thermodynamic system is its total mechanical energy.

Conceptual Questions

Is it possible to determine whether a change in internal energy is caused by heat transferred, by work performed, or by a combination of the 2?

When a liquid is vaporized, its change in internal energy is not equal to the oestrus added. Why?

Some of the energy goes into irresolute the phase of the liquid to gas.

Why does a bike pump feel warm as you inflate your tire?

Is it possible for the temperature of a organisation to remain constant when rut flows into or out of it? If and then, requite examples.

Yes, every bit long as the work done equals the heat added there will exist no alter in internal energy and thereby no change in temperature. When h2o freezes or when ice melts while removing or adding oestrus, respectively, the temperature remains constant.

Problems

A gas at a pressure level of 2.00 atm undergoes a quasi-static isobaric expansion from 3.00 to 5.00 L. How much work is done by the gas?

It takes 500 J of work to compress quasi-statically 0.50 mol of an ideal gas to one-fifth its original book. Calculate the temperature of the gas, assuming information technology remains constant during the pinch.

74 Thou

It is found that, when a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work. Assuming that the gas temperature remains abiding at 300 K, how many moles of gas are present?

In a quasi-static isobaric expansion, 500 J of work are done by the gas. If the gas pressure is 0.lxxx atm, what is the fractional increase in the volume of the gas, bold it was originally at 20.0 L?

1.four times

When a gas undergoes a quasi-static isobaric change in volume from 10.0 to ii.0 50, fifteen J of work from an external source are required. What is the force per unit area of the gas?

An ideal gas expands quasi-statically and isothermally from a state with pressure p and volume V to a state with book 4V. Show that the work washed by the gas in the expansion is pV(ln 4).

pVln(iv)

As shown below, calculate the piece of work done past the gas in the quasi-static processes represented past the paths (a) AB; (b) ADB; (c) ACB; and (d) ADCB.

(a) Calculate the work done by the gas along the closed path shown below. The curved section between R and S is semicircular. (b) If the process is carried out in the opposite direction, what is the work washed past the gas?

a. 160 J; b. –160 J

An ideal gas expands quasi-statically to 3 times its original volume. Which process requires more work from the gas, an isothermal process or an isobaric one? Determine the ratio of the work done in these processes.

A dilute gas at a pressure of 2.0 atm and a volume of 4.0 L is taken through the following quasi-static steps: (a) an isobaric expansion to a volume of ten.0 Fifty, (b) an isochoric alter to a pressure of 0.50 atm, (c) an isobaric pinch to a volume of four.0 L, and (d) an isochoric alter to a pressure of two.0 atm. Show these steps on a pV diagram and determine from your graph the net work done past the gas.

$W=900\phantom{\rule{0.2em}{0ex}}\text{J}$

What is the average mechanical free energy of the atoms of an ideal monatomic gas at 300 1000?

What is the internal energy of half-dozen.00 mol of an ideal monatomic gas at $200\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ?

$3.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{J}$

Summate the internal free energy of xv mg of helium at a temperature of $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$

Two monatomic platonic gases A and B are at the aforementioned temperature. If 1.0 g of gas A has the aforementioned internal energy as 0.10 g of gas B, what are (a) the ratio of the number of moles of each gas and (b) the ration of the diminutive masses of the ii gases?

a. 1:1; b. 10:one

The van der Waals coefficients for oxygen are $a=0.138\phantom{\rule{0.2em}{0ex}}\text{J}·{\text{m}}^{3}\text{/}{\text{mol}}^{2}$ and $b=3.18\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}{\phantom{\rule{0.2em}{0ex}}\text{m}}^{3}\text{/}\text{mol}$ . Use these values to draw a van der Waals isotherm of oxygen at 100 1000. On the same graph, describe isotherms of one mole of an ideal gas.

Find the piece of work done in the quasi-static processes shown below. The states are given equally (p, V) values for the points in the pV plane: 1 (3 atm, 4 L), 2 (iii atm, half dozen L), iii (v atm, iv 50), iv (two atm, 6 50), 5 (iv atm, 2 L), 6 (5 atm, 5 Fifty), and vii (two atm, five Fifty).

a. 600 J; b. 0; c. 500 J; d. 200 J; e. 800 J; f. 500 J

Glossary

internal energy: average of the full mechanical energy of all the molecules or entities in the system

quasi-static process: evolution of a system that goes so slowly that the arrangement involved is always in thermodynamic equilibrium

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Source: https://opentextbc.ca/universityphysicsv2openstax/chapter/work-heat-and-internal-energy/

what happens to the internal energy of a system when work is done on it

fifteen Work, Heat, and Internal Energy

Learning Objectives

Work Washed past a System

Internal Free energy

Summary

Conceptual Questions

Problems

Glossary

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