Topic 10: Thermal physics 10.2 Processes

The first law of thermodynamics

10.2.1 Deduce an expression for the work involved in a volume change of a gas at constant pressure.

10.2.2 State the first law of thermodynamics.

10.2.3 Identify the first law of thermodynamics as a statement of energy conservation.

10.2.4 Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.

10.2.5 Draw and annotate thermodynamic processes and cycles on P-V diagrams.

10.2.6 Calculate from a P-V diagram the work done in a thermodynamic cycle.

10.2.7 Solve problems involving state changes of a gas.

Topic 10: Thermal physics10.2 Processes


Deduce an expression for the work involved in a volume change of a gas at constant pressure.

Suppose we take a beaker that is filled with an ideal gas, and stopper it with a gas-tight cork and a weight, as shown.

The weight F causes a pressure in the gas having a value given by P = F/A, where A is the area of the cork in contact with the gas.

If we now heat up the gas it will expand against the cork, pushing it upward:

The dashed red box shows the change in volume ∆V.

Note that ∆V = Ax.


x

P is constant. Why?

∆VA


Deduce an expression for the work involved in a volume change of a gas at constant pressure.

From the previous slide: P = F/A F = PA and ∆V = Ax.The work W done by the gas is just the force F it exerts on the weighted cork times the distance x it moves the cork. Thus

W = Fx = PAx = P∆V.

FYIIf ∆V > 0 (gas expands) then W > 0.If ∆V < 0 (gas contracts) then W < 0.


x∆VA

F

Work done by expanding gas

(constant P)

W = P∆V


State the first law of thermodynamics.

Consider the previous “system” containing gas, and heat, and providing mechanical work.

Just as we have done before, we will use Q to represent a quantity of heat. If heat is added to our system (as it was) then Q > 0. If heat is removed from our system then Q < 0.

If the gas expands then W > 0. If the gas contracts then W < 0.

Finally, we define the change in internal energy of the system as ∆U.

The first law of thermodynamics relates Q, W and ∆U as follows:


first law of thermodynamics

Q = ∆U + W


Identify the first law of thermodynamics as a statement of energy conservation.

What the first law shows is that when heat energy Q is added to a system, some (or all of it) may be used to change the internal energy ∆U of the system, and some (or all of it) may be used to provide mechanical work W.



Q = ∆U + W

FYISometimes the first law is expressed ∆U = Q – W.In this form we see that there are two ways to change the internal energy of a system: (1)By passing heat Q through the system boundary.(2)By doing work W through the system boundary.


Identify the first law of thermodynamics as a statement of energy conservation.

In words, the first law says “Heat added to a closed system can change its internal energy and cause it to do work on its environment.”

This is a statement of the conservation of energy. All of the energy added to the system is accounted for.



Q = ∆U + W

FYIRecall that ∆U consists of both ∆EP (manifested in phase change) and ∆EK (temperature change) of the substance. Lack of internal forces in an ideal gas means that ∆EP = 0.


Describe the isochoric (isovolumetric), isobaric, isothermal, and adiabatic changes of state of an ideal gas.

For an ideal gas, the equation of state (PV = nRT) tells us all of the important things about a gas through the state variables P,V, n, and T.

When we add (or remove) heat Q from a closed system containing an ideal gas, or have it do work W on the external world, its state variables may change.

We call changing the state of a system a process.A process in which the state of a gas is changed without changing its volume is called isochoric.


FYIDon’t confuse “state” with “phase” in Topic 10.



A process in which the state of a gas is changed without changing its volume is called isochoric.

We have already seen an isochoric process when we studied the concept of absolute zero:


0

10 20

30

p

T (°C)-300 -200 -100 0 100 200 300

During an isochoric process the temperature and the pressure change.

NOT the volume.





PRACTICE: Show that the first law of thermodynamics reduces to Q = ∆U for an isochoric process.SOLUTION:Recall that the work done by a gas is given by W = P∆V.Isochoric means ∆V = 0. Thus W = 0.The first law, Q = ∆U + W, thus reduces toQ = ∆U.





EXAMPLE: Show that for an isolated ideal gas P T during an isochoric process.SOLUTION: Use PV = nRT. Then P = (nR/V)TIsolated means n is constant (no gas is added to or lost from the system).Then n and V are constant (as is R). Thus P = (nR/V)T = (CONST)T P T. (Isochoric)



A process in which the state of a gas is changed without changing its pressure is called isobaric.

We have already seen such a process, when we derived the formula for the work done by a gas:

For an isobaric process the first law can therefore be written

Q = ∆U + P∆V.


x∆VA

Work done by expanding gas

(constant P)

W = P∆V





EXAMPLE: Show that for an isolated ideal gas V T during an isobaric process.SOLUTION: Use PV = nRT. Then V = (nR/P)TIsolated means n is constant (no gas is added to or lost from the system).Then n and P are constant (as is R). Thus V = (nR/P)T = (CONST)T V T. (Isobaric)





PRACTICE: Show that for an isolated ideal gas W = nR∆T during an isobaric process.

SOLUTION:From PV = nRT we can write (if n and P are constant)

P∆V = nR∆T.Recall W = P∆V.Thus

W = nR∆T. (Isobaric)



If the state of a gas is changed without changing its temperature the process is called isothermal.


EXAMPLE: A graduated syringe which is filled with air is placed in an ice bath and allowed to reach the temperature of the water. Demonstrate that P1V1 = P2V2. SOLUTION:Record initial states after a wait:P1 = 15, V1 = 10, and T1 = 0ºC.Record final states after a wait:P2 = 30, V2 = 5, and T2 = 0ºC.P1V1 = 15(10) = 150 = 30(5) = P2V2.

0

10 20

30

How do we know that the process is isothermal?

Why do we wait before recording our values?



If the state of a gas is changed without changing its temperature the process is called isothermal.


PRACTICE: Show that for an isolated ideal gas P1V1 = P2V2 during an isothermal process.

SOLUTION:From PV = nRT we can write (if n and T are constant) P1V1 = nRT P2V2 = nRT.Thus

P1V1 = nRT = P2V2. (Isothermal)



If the state of a gas is changed without adding or losing heat the process is called adiabatic.


PRACTICE: Show that for an isolated ideal gas W = -∆U during an adiabatic process.SOLUTION:From Q = ∆U + W we can write (if n is constant and Q is zero) Q = ∆U + W 0 = ∆U + W W = -∆U.FYIWe accomplish this by insulating the container.


Draw and annotate thermodynamic processes and cycles on P-V diagrams.

Perhaps you have enjoyed the pleasures of analytic geometry and the graphing of surfaces in 3D.

The three variables of a surface are x, y, and z, and we can describe any surface using the "state" variables x, y, and z:

The equation "of state" of a sphere is x2 + y2 + z2 = r2, where r is the radius of the sphere:


FYIWe “built” the 3D sphere with layers of 2D circles.We have transformed a 3D surface into a stack of 2D surfaces.

y

z

x



The three state variables (if n is kept constant) of a gas are analogous.

We can plot the three variables P, V, and T on mutually perpen- dicular axes like this:

We have made layers in T. Thus each layer has a single temperature.


FYIEach layer is an isotherm.The 3D graph (below) can then be redrawn in its simpler 2D form (above) without loss of information.

V

P

T

V

P isotherms

T1T2

T3

T4



A thermodynamic process involves moving from one state to another state. This could involve changing any or even all of the state variables (P, V, or T).


EXAMPLE: In the P-V graph shown, identify each process type as ISOBARIC, ISOTHERMAL, OR ISOVOLUMETRIC (isochoric).SOLUTION: AB is isothermal (constant T).BC is isobaric (constant P).CA is isochoric (constant V).FYIThe purple line could be an adiabatic process.

P isotherm

isotherm

A

B

C

V



A thermodynamic cycle is a set of processes which ultimately return a gas to its original state.


EXAMPLE: A fixed quantity of a gas undergoes a cycle by changing between the following three states: State A: (P = 2 Pa, V = 10 m3) State B: (P = 8 Pa, V = 10 m3) State C: (P = 8 Pa, V = 25 m3)Each process is a straight line, and the cycle goes like this: ABCA. Sketch the complete cycle on a P-V diagram.SOLUTION:Scale your axes and plot your points…

P

V

2

8

10 25

A

B C


Calculate from a P-V diagram the work done in a thermodynamic cycle.



EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example):(a) Find the work done during the process AB.(b) Find the work done during the process BC.SOLUTION: Use W = P∆V.(a) From A to B: ∆V = 0. Thus the W = 0.(b) From B to C: ∆V = 25 – 10 = 15; P = 8. Thus W = P∆V = 8(15) = 120 J.

P

V

2

8

10 25

A

B C

EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example):(c) Find the work done during the process CA.SOLUTION: (c) Use W = Area under the P-V diagram.Observe that ∆V is negative when going from C (V = 25) to A (V = 10).Observe that P is NOT constant so W P∆V.W = Area = -[ (2)(15) + (1/2)(6)(15) ] = -75 J.





P

V

2

8

10 25

A

B C

EXAMPLE: A fixed quantity of a gas undergoes the cycle shown here (from the last example):(d) Find the work done during the cycle ABCA.SOLUTION: (d) Just total up the work done in each process.

WAB = 0 J.

WBC = +120 J.

WCA = -75 J.

Wcycle = 0 + 120 – 75 = +45 J.

FYIWork is done on the external environment during each cycle.





P

V

2

8

10 25

A

B C

PRACTICE: Find the total work done if the previous cycle is reversed.SOLUTION: We want ACBA.WAC

= Area

= +[ (2)(15) + (1/2)(6)(15) ]

= +75 J.

WCB = P∆V = 8(10–25) = -120 J.

WBA = 0 J (since ∆V = 0).

Wcycle = 75 - 120 + 0 = -45 J.

FYIReversing the cycle reverses the sign of the work.





P

V

2

8

10 25

A

B C



A heat engine is any device which converts heat energy into mechanical work.

To make a heat engine we need a source of heat energy (coal, gas, etc.) and a working fluid which undergoes thermodynamic change of state causing work do be done on the external environment.

Common working fluids are water (made into steam) and petrol-air mixtures (ignited).

The energy flow diagram of a heat engine is shown here:


hot reservoir at TH

cold reservoir at TL

engine

QH

QL

W

EXAMPLE: An internal combustion engine is an example of a heat engine that does work on the environment. A four-stroke engine is animated here.

http://www.animatedengines.com/otto.htmlhttp://chemcollective.org/activities/simulations/engine




http://www.animatedengines.com/otto.html

http://chemcollective.org/activities/simulations/engine



EXAMPLE: From the animation of the second link on the previous slide, show the direction of each process with an arrow.

SOLUTION:View the animation. Or reason that POSITIVE work must be done by the cycle.FYIRemember “work done BY system” is positive, and “work done ON system” is negative.




EXAMPLE: From the same animation, label the COMPRESSION, POWER, EXHAUST, AND INTAKE STROKES. Which stroke has a positive area under it? Which has a negative area? SOLUTION:A STROKE involves a change in volume.Since ∆V > 0 for the power stroke it has a positive area.The compression stroke has a negative area.




COMPRESSION STROKE

POWER STROKE

EXHAUST STROKE

INTAKE STROKE

EXAMPLE: From the same animation, show that the work done in a cycle is positive.SOLUTION:The work done during the power stroke is positive (sketched in red).The work done during the compression stroke is negative (sketched in purple).There is more positive than negative. Thus W > 0.W = area between graphs.




COMPRESSION STROKE

POWER STROKE

EXHAUST STROKE

INTAKE STROKE



A heat pump is any device which can move heat from a low temperature reservoir to a high temperature reservoir.

A refrigerator is an example of a heat pump.

A common working fluid is Freon.The Freon vaporizes inside the fridge removing QL = LV.

The Freon condenses outside the fridge releasing QH = Lf.


hot reservoir at TH

cold reservoir at TL

pump

QH

QL

W

FYIThe compressor does the work ON the system.


Solve problems involving state changes of a gas.


”Thermally insulated” means that heat can not enter or leave the system.

Thus Q = 0.

This is the definition of an adiabatic process.




W = P∆V= 1105(-3)

= -3106 J.

Or you can just find the area under the process.




Adiabatic (and insulated) both mean that Q = 0.

The first law says Q = ∆U + W. Then

∆V < 0 means W < 0 (-).

0 = ∆U + (-) ∆U > 0.




Isothermal means T is constant.PV = nRT then becomes PV = CONST.

At A: PV = 25 = 10.At B: PV = 52 = 10.At C: PV = 6.82 = 13.6.

Thus process AB is the isothermal one.




The difference in work is the area between the graphs.




This can only be an estimate.A tally of small rectangles should suffice.

18161412121099887765544

19There are about 170 of them.

Each rectangle has a work value given byW = PV = (0.1105)(0.110-3) = 1 J.

Thus the difference in work is 170 J.




The first law saysQ = ∆U + W.

Adiabatic means Q = 0.∆U = -W.

AC is the adiabatic compression.

Since W < 0 during the compression of a gas, ∆U = -W > 0. Thus the temperature increases.




The gas only does external work when it expands.QR and SP are isochoric (isovolumetric) so W = 0.PQ is a compression so W < 0.RS is an expansion so W > 0.




NOT adiabatic since Q 0 (Q = 8103 J).From PV = nRT we see that

PiVi = 1.2105(0.05) = 6000 J = nRTi.PfVf = 1.2105(0.10) = 12000 J = nRTf.

Thus T changes and the process is NOT isothermal.So the process is NEITHER.




From W = P∆V we see that

W = 1.2105(0.10 - 0.05) = 6.0103 J.




From the first law: Q = ∆U + W, or

8.0103 = ∆U + 6.0103

∆U = 2.0103 J.




Work done ON a gas is a compression. Only process PQ has a decreasing volume.




For an ideal gas, ISOTHERMAL means ∆U = 0. From the first law of thermodynamics we have

Q = ∆U + WQ = 0 + WQ = W = 2500 J.




From the first law we have

q = ∆U + w.

Keep T constant with ICE BATH.

Keep V constant with FIXED CONTAINER.

Keep q zero with INSULATION.




A heat pump transfers heat from a cold reservoir to a hot one.

QH

QL

Since this is not the natural direction of flow, work must be done ON the system to make this happen.

W




It is clear from the diagram that both AB and CD are isobaric (P = CONST).

From CD evaporation occurs. During phase changes ∆T = CONST (isothermal).From AB compression occurs. During phase changes ∆T = CONST (isothermal).




Wherever W > 0 heat is removed from the cold reservoir. Thus BC and CD.

Wherever W < 0 heat is forced into the hot reservoir. Thus DA and AB.




From the diagram Ein = W + Qc. Eout = QH.

From Ein = Eout we have W + QC = QH.




A refrigerator is a heat pump. Work must be done ON it. Thus W < 0. Area inside curves must be NEGATIVE.

P = CONST

A

A

B

Wherever W > 0 heat is removed from the cold reservoir. Thus where ∆V > 0.

B




For each square:

W = P∆V

W = (1105)(0.110-3)

W = 10 J.

There are about 47 squares.

Thus W = 47(10 J) = 470 J (done ON substance).

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Topic 10: Thermal physics 10.2 Processes