Thermodynamics

The first law of thermodynamics

Heat is a form of energy.  It is disordered energy.  Work is the conversion of one form of energy into another form.
Can we do work with heat?  Can we convert it into another form of energy? 
Thermodynamics is the study of the way that one does work with heat.
A very simple device that can convert heat into potential energy is a rubber band.  Unlike most other substances, rubber contracts when heated.  We can therefore lift an object by heating a rubber band.  Heat is converted into gravitational potential energy.

A simple rubber band can be the heart of an engine converting heat into electrical energy.  We can use heat to make the rubber band contract.  It then can lift water, converting heat into gravitational potential energy.  This gravitational potential energy of the water can be converted into electrical energy with the help of a turbine generator.  After the load had been lifted, we let the rubber band cool down.  It then has the same internal energy it had before we started the process.  We can repeat this process over and over again.  This certainly is a very inefficient engine.  It wastes a large amount of heat.  Can we make the process of converting heat into other forms of energy more efficient?

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How much work can we possible get out of heat? 

Energy conservation limits the amount of work we can get out of a certain amount of heat. 

The first law of thermodynamics state that energy is conserved. 

We may express it in the following way:

Increase in internal energy of a system

= heat put into the system + work done on the system by its surroundings,
or
ΔU = ΔQ + ΔW
A system can be anything.  It is most convenient if it has well defined boundaries.
ΔQ is positive if heat flows into the system, negative if it flows out of the system. 
ΔW is positive if work is done on the system and it is negative if the system does work on its surroundings. 
The internal energy is the sum of the kinetic and potential energy of the atoms and molecules that make up the system.

Problem:
Assume you are running an electric space heater.  Let the space heater be the system under consideration.  It has warmed up and is now running at a constant temperature.  It consumes 500 Watts of electrical power.  How much electrical energy is converted into heat per hour?  (1 Watt = 1 J/s)

Solution:
ΔU = ΔQ + ΔW. 
ΔU = 0 since the temperature of the heating element is constant.
Therefore ΔQ = -ΔW. 
ΔW is the work done by the power supply in providing electrical energy to the system.  It is positive.  ΔQ must be negative.  It is the heat leaving the system.  500 J of heat are leaving the system per second. 
500 J/s * 3600 s = 1.8 MJ of electrical energy are converted into heat every hour.  All the electrical energy put into the system is converted into heat.  (Any form of ordered energy can be completely converted into heat.)

Can we reverse the process?  Can we take the heat generated by the space heater or by some other process and completely convert it back to electrical energy?  Can we design an engine (a device) that accomplices this?  The first law of thermodynamics does not exclude this possibility.  It just says that we will not get more electrical energy out of a system than the energy we put into the system in the form of heat, assuming ΔU = 0, i.e. the internal energy of the system is the same before and after doing the work.

Often the system from which we want to extract heat to do work is a gas.  When a gas expands, how much work does it do on its surroundings?

Consider an ideal gas in a cylinder with a movable piston.  The gas occupies a volume V, the temperature is T, and the pressure is P.  Assume that the piston has a cross-sectional area A.  Assume the gas expands by a small amount and the piston moves out by a small amount dy.  The gas exerts a force F = PA on the piston and therefore does work dW = Fdy on the piston.

dW = Fdy = PAdy = PdV

dV is the change in volume of the gas.  dV is positive because the gas expands.  (If the gas is being compresses, then dV is negative and work is done on the gas.)

If the volume of the gas changes from V1 to V2, then the total work done by the gas is

.

Note:  Here W is the work done by the gas, not the work done on the gas.

This work depends on how exactly the pressure varies during the expansion process.  If the pressure and volume are known at each step of the expansion process, then the work can be represented as the area under the curve on a PV diagram.

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An adiabatic process is a process during which no heat enters or leaves the system.  We then have DU = -DW.  In practice there are two different ways to prevent the transfer of heat.
(a) Provide very good thermal insulation of the system.
(b) Complete the process in a very short time interval, so  that there is no time for appreciable heat transfer.  The combustion process inside a car engine is essentially adiabatic for this reason.
An isobaric process is a process that occurs at constant pressure.  We then have

.

If the pressure of an ideal gas is kept constant, then the temperature must increase as the gas expands.  (PV/T = constant.)  Heat must be added during the expansion process.

An isovolumetric or isometric process takes place at constant volume.  Then W = 0 and DU = DQ.  All the heat added to the system goes into increasing its internal energy.
An isothermal process occurs at constant temperature.  Since the internal energy of a gas is only a function of its temperature, DU = 0 for an isothermal process.  For the isothermal expansion of an ideal gas we have

.

W is positive if V2 > V1.  Since DU = 0, the heat transferred to the gas is DQ = W.

Problems:

Gas in a container is at a pressure of 1.5 atm and a volume of 4 m3.   What is the work done by the gas if
(a) it expands at constant pressure to twice its initial volume?
(b) it is compressed at constant pressure to 1/4 of its initial volume?
Solution:
(a) W = P(V2-V1) = P(2V1-V1) = PV1 = 151500 Pa´4 m3 = 606000 J
(b) W = P(0.25V1 - V1) = -0.75PV1 = -454500 J
A gas is compressed at a constant pressure of 0.8 atm from 9 L to 2 L.   In the process, 400 J of thermal energy leaves the gas.
(a) What is the work done by the gas?
(b) What is the change in its internal energy?
Solution:
(a)  W = P(V2-V1) = -80800 Pa ´ 7´10-3 m3 = -565.6 J
(b)  DU = -400J + 565.6J = 165.6J
A thermodynamic system undergoes a process in which its internal energy decreases by 500 J.  If at the same time 220 J of work is done on the system, find the thermal energy transferred to or from it.
Solution:
increase in internal energy of a system
= heat put into the system - work done by the system on its surroundings,
or
DU = DQ - DW.
-500J = DQ + 220 J
DQ = -720 J
One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1 atm and volume of 25 L.  Determine
(a)  the initial volume and
(b) the temperature of the gas.
Solution:
(a)  For an isothermal process the temperature is constant.  
Therefore PV = nRT = constant.
PV = 101000 Pa´25´10-3 m3 = nRT
for an isothermal process.
W/(nRT) = 3000 J/(101000 Pa´25´10-3 m3) = 1.19
V2/V1 = exp(1.19) = 3.28
V1 = (25´10-3 m3)/3.28 = 7.62´10-3 m3 = 7.62 L
(b) For an ideal gas PV = nRT. 101000Pa´25´10-3 m3 = (8.31 J/K)T. T = 303.85K.

A bicycle ergometer is an apparatus that resembles a bicycle and is used to measure the amount of work done by a person while pedaling to rotate a large flywheel, usually the front wheel of a stationary bicycle, against a frictional force.  The work done to overcome the frictional force is converted into thermal energy, which causes the temperature of the flywheel and of other components in contact with the system to increase.  A nylon belt, which is under tension and wound around the wheel, can provide the frictional force.  The tension can be measured with a calibrated spring.  The work W done by the person on the ergometer is the product of the force due to the tension F in the belt times the distance through which this force acts.  The force is tangential to the rim of the wheel.  The distance is therefore 2p times the radius of the flywheel r times the number of revolution N made by the wheel.  We have W = NF2pr.  For the ergometer we have from the first law of thermodynamics DU = DQ - DW.  DW is positive if the system does work on its surroundings and is negative if work is done on the system.  Here DW is negative, DW = -W.  DQ is the heat put into or taken out of the system.  We have DU-DQ = NF2pr.  If we neglect the heat lost by the system to its surroundings and set DQ = 0, then the increase in the thermal energy of the system equals DU = NF2pr.  The temperature of the flywheel therefore increases by DT = DU/mc, where m is the mass of the flywheel and c is its specific heat capacity. 
The ergometer converts useful work into thermal energy.  By analyzing the air exhaled by the person pedaling the ergometer, the rate at which the body uses chemical energy can be measured.  By knowing how much chemical energy the body uses at rest and how much it uses while pedaling, the efficiency of the body in converting chemical energy into useful work can be determined.

The second law of thermodynamics

One of the earliest scientists to be intrigued by heat engines was a French engineer named Sadi Carnot (1796-1832.  He assumed that an ideal engine would be a frictionless engine.  It would also be a reversible engine.  By itself, heat always flows from an object of higher temperature to an object with lower temperature.  A reversible engine is an engine in which the heat transfer can change direction, if the temperature of one of the objects is changed by a tiny (infinitesimal) amount.  When a reversible engine causes heat to flow into a system, it flows as the result of infinitesimally small temperature differences, or because there is an infinitesimal amount of work done on the system. If such a process could be actually realized, it would be characterized by a continuous state of equilibrium (i.e. no pressure or temperature differentials) and would occur at a rate so slow as to require an infinite time.  A real engine always involves at least a small amount of irreversibility.  Heat will not flow without a temperature differential and friction cannot be entirely eliminated.

Carnot showed, that if an any real reversible engine picks up the amount of heat Q1 from a reservoir at temperature T1, converts some of it into useful work, and delivers the amount of heat Q2 to a reservoir at temperature T2, then Q1/T1 = Q2/T2.  Here T is the absolute temperature, measured in Kelvin and a heat reservoir is a system, such as a lake, that is so large that its temperature does not change when the heat involved in the process considered flows into or out of the reservoir.  To convert heat into work, you need at least two places with different temperatures.  If you take in Q1 at temperature T1 you must dump at least Q2 at temperature T2

The useful work done by a heat engine is W = Q1 - Q2 (energy conservation).  An ideal reversible engine does the maximum amount of work.  Any real engine delivers more heat Q2 at the reservoir at T2 than a reversible one and therefore does less useful work.  The maximum amount of work you can therefore get out of a heat engine is

Wmax = Q1 - Q2 = Q1 - Q1 T2 / T1 = Q1 (1 – T2 / T1).

W is positive if T1 is greater than T2. 

Assume you have a reservoir of hot water at temperature T1.  Can you take an amount of heat Q1 out of this reservoir and convert it into work?  No!  You can convert a fraction of the heat into work if you have a place at a lower temperature T2 where you can dump some of the heat.  An engine that does work by removing heat from a reservoir at a single temperature can not exist.  Heat cannot be taken in at a certain temperature with no other change in the system and converted into work.  This is one way of stating the second law of thermodynamics.

Heat cannot, of itself, flow from a cold to a hot object is another way of stating the second law of thermodynamics.  If it could, then heat dumped at T2 could just flow back to the reservoir at T1 and the net effect would be an amount of heat ΔQ = Q1 - Q2 taken at a T1 and converted into heat with no other changes in the system.

Reversible engines

Let us consider an idealized, frictionless engine in which all the processes are reversible.  To show that such an engine is possible in principle, if not in practice, let us consider an example.  Assume that we have an ideal gas in a cylinder equipped with a frictionless piston.  Also assume that we have two heat reservoirs at temperatures T1 and T2, with T1 higher than T2.  

  1. Let us first put the cylinder in contact with the reservoir at T1 and heat the gas and at the same time expand it.  To make the process reversible we pull the piston out very slowly as heat flows into the gas and we make sure that the temperature of the gas stays nearly equal to T1.  If we would push the piston back in slowly, then the temperature would only be infinitesimally higher than T1 and the heat would flow back from the gas into the reservoir.  An isothermal expansion, when done slowly enough, can be a reversible process.

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    The figure above shows a plot of the pressure of the gas vs. its volume.  As the gas expands, the pressure falls.  The curve marked (1) from point a to point b tells us how the pressure and volume change if the temperature is kept fixed at the value T1.  For an ideal gas this curve is given by 

    PV = NkT1.

    Once we reach point b in the figure an amount of heat Q1 has been transferred from the reservoir into the gas.  Since the expansion is isothermal and the temperature of the gas has not changed, DU = 0 and Q1 = DW.

    .

  2. Let us take the cylinder away from the reservoir at point b and continue a slow, reversible expansion without permitting heat to enter the cylinder.  The expansion now is adiabatic.  As the gas expands the temperature falls, since there is no heat entering the cylinder.  We let the gas expand, following the curve marked (2), until the temperature falls to T2 at the point marked c.  The adiabatic curve has a more negative slope than the isothermal curve.  For the adiabatic expansion of an ideal gas made from point particles, we have DU = -DW since DQ = 0. Therefore

    dU = -dW = -PdV.

    But we also have 

    U = N(1/2)m<v2>,

    since the internal energy is the random kinetic energy of the gas atoms. Using, from the kinetic theory, PV = (2/3)N(1/2)m<v2> we have 

    U = (3/2)PV. 

    Then, from calculus, we have 

    dU = (3/2)(PdV + VdP). 

    Equating our two expressions for dU we have 

    -PdV = (3/2)(PdV + VdP), 
    (-5/2)PdV = (3/2)VdP, 
    dP/P + (5/3)dV/V = 0. 

    We can integrate to obtain lnP + (5/3)lnV = lnC, where lnC is the constant of integration. This yields

    PV5/3 = C = constant 

    for an adiabatic expansion of an ideal gas.  Since PV = NkT we can write PV5/3 = NkTV2/3 = C, or TV2/3 = constant.  Therefore 

    T1Vb2/3 = T2Vc2/3  or  Vb/Vc = (T2/T1)-2/3.

  3. When the gas has reached the temperature T2 we put it in contact with the reservoir at T2.  Now we slowly compress the gas isothermally while it is in contact with the reservoir at T2, following the curve marked (3).  The temperature of the gas does not rise and an amount of heat Q2 flows from the cylinder into the reservoir at the temperature T2.  For the isothermal process we have

    .

  4. At the point d we remove the cylinder from the reservoir at T2 and compress it still further, without letting any heat flow out.  During this adiabatic process the temperature rises, and the pressure follows the curve marked (4).  If we carry out each step properly, we can return to the point a at temperature T1 where we started, and repeat the cycle.

    For the adiabatic process from point d to point a we have

    T2Vd2/3 = T1Va2/3  or  Va/Vd = (T2/T1)-2/3.

    Therefore

    Va/Vd = Vb/Vc,   Va/Vb = Vd/Vc,  and  ln(Va/Vb) = ln(Vd/Vc).

In going from point a to point c, the gas expands and does work.  In going from point c back to point a work is done on the gas.  The work done is always the area under the curve in the PV diagram.  The net amount of work done by the gas is the yellow area of the figure.  Because everything we have done is reversible, we could also have gone backwards instead of forwards through the cycle.  Then the area under the curve would represent the net amount of work done on the gas.

During one cycle we have put an amount of heat Q1 into the gas at temperature T1 and have removed an amount of heat Q2 at temperature T2.  From the above equations we see that

Q1/Q2 = T1/T2, or Q1/T1 = Q2/T2.

We have shown that if our ideal reversible engine picks up the amount of heat Q1 from a reservoir at temperature T1 and delivers the amount of heat Q2 to a reservoir at temperature T2, then Q1/T1 = Q2/T2.  The work done by the engine on its surroundings is Q1 - Q2.

Let us now assume that we have another engine which takes Q1 at T1, does work W, and delivers some heat at T2.  If W were greater than the work W done by our reversible engine, then we could use W to run our reversible engine backwards and put Q1 back into the reservoir at T1. Then all that we would have effectively done by running both engines is to take heat out of the reservoir at T2 and completely converted it into the useful work W - W.  But according to the second law of thermodynamics it is impossible to obtain useful work from a reservoir at a single temperature with no other changes.  Heat cannot be taken in at a certain temperature with no other change in the system and converted into work.  Therefore no engine which absorbs a given amount of heat from a higher temperature T1 and delivers it at the lower temperature T2 can do more work than a reversible engine operating under the same temperature conditions.  If our second engine is also reversible, then W must be equal to W, or we could reverse the above argument.  If both engines are reversible they must both do the same amount of work, and Q1/T1 = Q2/T2 for both engines.  If an engine is reversible, it makes no difference how it is designed.  The amount of work it does when it absorbs a given amount of heat at temperature T1 and delivers heat at some other temperature T2 does not depend on the design of the engine.  It is a property of the world, not a property of a particular engine.

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