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# Expansion and Compression Work

last edited by 10 years, 11 months ago

Expansion and Compression Work

Introduction

In thermodynamics, one of the primary methods to find the value of a system’s properties, such as its temperature, pressure, and volume, is to find the work that is experienced by the system. Work is measured by the expansion or contraction of the system, and when work is added to the thermal energy of the system, the total internal energy of the system can be found. For the sake of simplicity, this exploration will consider the expansion-contraction work of ideal gases, as they  vastly simplify the relationships between pressure, temperature, and volume of a system.

Ideal Gases

An ideal gas is defined as a gas that fits the following rules:

1.      Intermolecular forces, forces between molecules, ions, and atoms, are inexistent.
2.      The volumes of the molecules are negligible compared to the volume of the container.
3.      Molecules constantly move in straight motions in various random directions.
4.      The molecular collisions are perfectly elastic. Thus, Kinetic Energy is retained in these collisions, and average kinetic energy is proportional to the temperature of the gas.

An ideal gas can be defined by the ideal gas law, which is

PV = nRT

P = Pressure in atmospheres (atm)    V=Volume in liters (L)   n= # of moles   R=.08206 (L*atm)/(mol*k)   T=Temperature in Kelvin

There is no such thing as an “ideal” gas. However a gas that behaves close to the rules that define what an ideal gas is and a gas that is close to Standard Temperature Pressure (STP), will approximately follow the ideal gas law. Standard Temperature Pressure is 273 Kelvin (zero degrees Celsius) and 1 atm (760 torr). For example, air is not an ideal gas, because it is made up of different molecules which can have intermolecular forces between one another, however, it still follows the ideal gas law, because it is at STP. Gases are least ideal at high pressures and low temperatures.

We must consider if gases are ideal in compression and expansions, because when we relate to the work that allows the compression or expansion, we are suggesting that the energy to do the work comes from a change in pressure and volume that is equal to the change in temperature, average kinetic energy, times the ideal gas constant, R, and the number of moles. If the gas was not ideal, we cannot make this assumption for the work done in expansion and compression.

Work of an Expanding or Contracting Gas

When it comes to measuring the work of an ideal gas that is expanding or contracting, work is equal to the negative of the change in pressure and volume. This is analogous to finding the work of a force in physics, where work is equal to force multiplied by displacement. For gases, pressure can be thought of as a force and the change in volume can be thought of as the displacement. Thus, if one wishes to find the work of a system under constant pressure, then work is:

Where Pext is the pressure experienced by the system, in most cases this is the external pressure of the system, and V represents volume. Note, P is measured in Pascals and volume is measure in meters cubed to obtain the work in units of joules.

If one wanted to find the work of a system under constant pressure, say 5 Pascals, and the volume change is 0.5 m3, then the work done by the system is 5 multiplied by 0.5, which is 2.5 Joules.

Thus, if one plots pressure versus volume, the work done, with constant pressure, is the area of the rectangle defined by V1 and V2 and the pressure. Thus, it is a rectangle of length V2-V1 and a height of P. See the graph below for reference.

The area of “W” is the work experienced by the system.

The above equations, and the ones below, only apply when the systems under study are quasi-static. A quasi-static system is one that is capable of reaching equilibrium, which means its properties, such as temperature, volume, and pressure, are capable of reaching a constant value. Even if a quasi-static system is perturbed by an external influence, such as heat, that changes one of its properties from equilibrium, the quasi-static system will eventually reach a new equilibrium, which makes quasi-static systems easy to model. Moreover, if a system were not at equilibrium or incapable of reaching equilibrium, then these systems do not have constant values for each of its properties, (ex: temperature, volume, and pressure). In addition, not all parts of the system are homogeneous if the system is not at equilibrium. For instance a certain area of the system could have a higher temperature than in other areas, making it impossible to characterize temperature for the entire system. Therefore, it is important to consider quasi-static systems for expansion and compression work as the equations in this exploration all rely on systems where a specific value of its properties for the entire system can be found.

Adiabatic and Isothermal Expansion/Compression of a Gas

The first process to be considered for compression work on ideal gases is the isothermal process.  When compression work is done under conditions where temperature remains constant, the process is said to be isothermal.  These constant temperature conditions are usually achieved by compressing the gas in the system at a rate such that the work done on the system equals the amount of heat being released during compression. To understand why this works to keep the temperature constant, recall the First Law of Thermodynamics where U is internal energy, q is heat, and w is work:

Note that the sign conventions to be used in this exploration are that q>0 for heat absorbed by the system, q<0 for heat released by the system, w>0 for work done on the system, and w<0 for work done by the system.

In our constant temperature system, although work is added to the system, an equal amount of heat is simultaneously being released from it, so overall, there is no change in internal energy.  For n=number of moles of gas, cv=molar heat capacity at constant volume, and T=temperature, the internal energy, U, of an ideal gas is given as:

Thus, the internal energy of an ideal gas depends only on temperature.  If there is no change in internal energy, then there is also no change in temperature and the process is indeed an isothermal one.

For isothermal processes, we have concluded that the change in internal energy is 0 so plugging this value into the First Law of Thermodynamics:

The work term, w, for compression is pressure-volume work.  For P= pressure and V=volume, pressure-volume work is given by:

Since the system is not held at a constant pressure, as the volume changes, pressure changes as well, so to get the total work, we need to account for small changes in pressure due to small changes in volume.  In other words, by looking at a pressure-volume graph for isothermal processes, it is seen that for a change in pressure, there is also a change in volume:

In order to fully calculate each of these small changes, we must use calculus to calculate the total area under the pressure-volume curve and we integrate the equation:

From the ideal gas law, PV=nRT, we know that:

So plugging this value into the integral:

Since n and R are constants and temperature is constant for isothermal processes, all of these variables are constant and can be taken out of the integral:

Taking the integral of dV/V:

Finally, applying a property of logarithms, ln(a)-ln(b)=ln(a/b):

For compression work, Vf < Vi, so w > 0 and this is the equation describing work done on the system.  Plugging this value in for w to the First Law:

This equation expresses the heat that flows out of the system.

Generally, isothermal processes occur very slowly so that each bit of work done on the system is then released as heat and the temperature of the gas doesn’t change.  If work was done on the system very quickly, then there will be a moment when the system is greatly heated as no heat can escape from the gas.  This kind of process, the adiabatic process, will be considered next.

In regards to an adiabatic process, there is no energy transferred between a system and its surroundings. This process is deemed to be “thermodynamically isolated,” which means that there is no heat transfer with the surroundings. An adiabatic process can happen if a system is encased by an insulated wall that’s very resistant to heat exchange or if the process occurs quickly such as in a compression. The adiabatic compression is usually a compression of a system that occurs extremely quickly so that there is no escape of heat from the gas.

Recall that the equation for internal energy, and also the first law is:

In an adiabatic process, there is no change in heat, so q = 0.

Thus, any change in internal energy is equivalent to work:

Remember that:

So:

For an ideal gas:

Then:

This means that when the gas expands under adiabatic conditions, it does work and loses energy. That is why its temperature drops.

In conditions where compression is done under constant temperatures, an isothermal process, there is energy being released as compression occurs. Not only is the system releasing energy in the form of heat, there is also work being done on the system by the surroundings, to keep it at the same temperature.

This is fundamentally different than an adiabatic system, as in an adiabatic system, no heat leaves or enters the system.

For example, if you have an equation xy=1 and x=1/2, then you can calculate that y=2, or if x=1/3, then y=3.  On the other hand, if the equation was xy2=1, if x=1/2, then y=sqrt(2) and if x=1/3, then y=sqrt(3). If you try to multiply x and y without the power, the result would not be constant since 2*sqrt(2)=2.828 and 3*sqrt(3)=5.196 are not the same.  But if you use the power, then (1/2)*((sqrt(2))2)=(1/3)((sqrt(3))2)=1 and this is constant.

γ is the adiabatic index, also called a heat capacity ratio.

where cp is the heat capacity for constant pressure and cv is the heat capacity for constant volume. Heat capacity is a value that determines the amount of heat required to change the temperature of a substance. For instance, the heat capacity of liquid water at constant pressure is about 4.18 J K-1mol-1. Thus, for one mole of water, it takes 4.18 Joules of energy to raise its temperature by one Kelvin. Moreover, this relationship can be used to find the energy given off or absorbed by water in a given temperature range. For instance, heating one mole of water from 10 to 20 Kelvin requires 4.18 J K-1mol-1 * 1 mol water * (20 Kelvin - 10 Kelvin) = 4.18*10 = 41.8 Joules.

For an ideal monoatomic gas, c­v = (3/2)R; note that cp = cv + R.

Those who are curious about howwas obtained, refer to the derivation below:

This causes the equations for isothermal and adiabatic processes to differ.

The first law of thermodynamics state that:

In isothermal processes, ΔU=0 since there's no change in temperature so:

However, in an adiabatic process, there is no change in q, or heat, which means that q =0.

So, any change in internal energy is purely based on work, ΔU = w.

Application of Concepts:

1. Energy and Work of a Piston

Imagine a gas in a piston that has a pressure and a volume denoted by A, B, C, and D. It is undergoing a cyclic process of changing its pressure and volume, which is described in the above graph. At which of those points is the piston moving and the gas performing work?

Note that at points A and C, volume is constant, and pressure is changing. Since volume is constant, then the work done by the gas is zero. Moreover, since volume does not change, then the piston does not move. This makes intuitive sense since work is zero at these points.

At points B and D, the gas is held at constant pressure, and it is doing work since volume is changing. At point B, volume is increasing, therefore the change in volume is positive. This means the gas is doing negative work. At point D, the volume is decreasing, therefore, the change in volume is negative. Thus, the gas does positive work at point D.

So what does negative and positive work mean? At point B, the gas is performing negative work since it is expanding. This means the gas is losing its ability to do further work, as the gas can only expand so much. At point C, the gas is contracting, and work is positive. This means the gas is gaining the ability to do work, as well as gaining the ability to do work by contracting.

If we assume that the thermal energy of the system is constant during the entire cyclic process, then the total change in internal energy of the system is dependent on the work done by the gas. (See the equation below)

Thus, at point D, the gas is gaining energy due to positive work but at point B, the gas is losing energy due to negative work. At points A and C, the gas has a constant energy since no work is done at these points. This also makes intuitive sense, since at B, the gas is releasing energy to its surroundings by expanding and doing work, while at point D, the gas is absorbing energy from its surroundings, gaining the ability to do work.

As seen in the graph above, an isothermal and an adiabatic process are not equivalent. In an isothermal process, pressure multiplied by volume is always constant. This is because of the relationship below:

Note that n, R, and T are constants, thus, PV=nRT=constant. In this case, temperature is constant because, by definition of an isothermal process, the expansion or compression occurs at a constant temperature. However, in an adiabatic process the product of pressure and volume is not constant; only the product of pressure and volume raised to a power γ (the adiabatic index or heat capacity ratio) is constant, as shown by the equation below:

Note that if it were not for γ, then this relationship would not be constant.

Thus, for an isothermal process, pressure is inversely proportional to volume, while in an adiabatic process, pressure is inversely to volume raised to a power. For an ideal, monatomic gas, that power, or γ, is 5/3. Therefore the function of pressure versus volume for an adiabatic process decays faster than the same function for an isothermal process.

3. Rising Bubbles

Imagine the following scenario:

“Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly so that no heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface?” - problem 1.38 in Thermal Physics, Daniel Schroeder.

Bubbles

First, bubble A is adiabatic, since no heat is exchanged with its environment and bubble B is isothermal since it exchanges heat with its environment. As proven above in #2, the adiabatic bubble will have a smaller volume than the isothermal bubble. The function of pressure versus volume for an adiabatic process decays faster than the same function for an isothermal process. Since the change in the external pressure experienced by both bubbles is equal, the adiabatic bubble will have a smaller volume while the isothermal bubble will have a larger volume. If this is confusing, consult the graph below:

Questions:

1. Are there any examples of an ideal gas in the real world?
2. If there is no change in the volume of a system, is there work being done?
3. If an isothermal process does 350 kJ of work, how much heat was absorbed by the system?

1. Ideal gases do not exist, however, they are a useful model to predict the behavior of non-ideal gases.
2. The most basic definition of work of a gas is that work is equal to pressure multiplied by the change in volume. If there is no change in volume, there is no work.
3. In an isothermal system, the work done is the opposite of the heat absorbed. Thus, if the system did 350 kJ of work, it absorbed 350 kJ of heat. However, if the system had 350 kJ of work done on it, it would have given off 350 kJ of heat.

References:

http://en.wikipedia.org/wiki/File:Isobaric_process.png

http://www.chemguide.co.uk/physical/kt/idealgases.html

http://web.mit.edu/16.unified/www/FALL/thermodynamics/thermo_2.htm

http://lorien.ncl.ac.uk/ming/webnotes/Therm1/revers/isothe.htm

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Thermal/ImportantThermalProcess.html

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html

An Introduction to Thermal Physics