Assignment 9: First and Second Laws of ThermodynamicsDue: 2:00am on Saturday, November 13, 2010

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A Flexible Balloon

A flexible balloon contains 0.320 of hydrogen sulfide gas . Initially the balloon of has a

volume of 7000 and a temperature of 29.0 . The first expands isobarically until the volume

doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the may be treated as an ideal gas with and .

Part A

What is the total heat supplied to the gas in the process?

Hint A.1 Which parts of the process have heat flow?

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Hint A.2 Relation of heat capacity and heat

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Hint A.3 Find the change in temperature

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ANSWER: = 3350

Correct

Part B

What is the total change in the internal energy of the gas?

Hint B.1 Dependence of internal energy on temperature

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ANSWER: = 0

Correct

Part C

What is the total work done by the gas?

Hint C.1 How to approach the problem

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ANSWER: = 3350

Correct

Part D

What is the final volume ?

Hint D.1 Relation between volume and temperature

Recall that, in an adiabatic process with finite changes in volume and temperature, is a

constant. As a result, .

ANSWER: = 0.112

Correct

Work from an Adiabatic Expansion

In this problem you are to consider an adiabatic expansion of an ideal diatomic gas, which means thatthe gas expands with no addition or subtraction of heat.This applet shows the adiabatic compression and expansion of an ideal monatomic gas with . It

will help you to see the qualitative behavior of adiabatic expansions, though your actual calculations willuse a slightly different .

Assume that the gas is initially at pressure , volume , and temperature . In addition, assume that

the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio ofheat capacities is .

Note that, unless explicitly stated, the variable should not appear in your answers--if needed use the

fact that for an ideal diatomic gas.

Part A

Find an analytic expression for , the pressure as a function of volume, during the adiabatic

expansion.

Hint A.1 Find the conserved quantity in an adiabatic process

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Express the pressure in terms of and any or all of the given initial values , , and .

ANSWER:

=

Correct

The fact that is a constant derives from the definition of an adiabatic process as one in

which no heat flow into or out of the system occurs. Setting in the first law of

which no heat flow into or out of the system occurs. Setting in the first law of

thermodynamics ( ) gives the starting point for this derivation: . See

your textbook for more details.

Part B

At the end of the adiabatic expansion, the gas fills a new volume , where . Find , the

work done by the gas on the container during the expansion.

Hint B.1 How to approach the problem

The work done is given by the following general integral:

.

Using the pressure as a function of volume, , found in Part A, integrate to find the total work

done as the gas expands from its initial volume to its final volume

Hint B.2 Help with the math

Integration gives

.

Express the work in terms of , , and . Your answer should not depend on temperature.

ANSWER:

=

Correct

Part C

Find , the change of internal energy of the gas during the adiabatic expansion from volume to

volume .

Hint C.1 Find the initial internal energy

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Hint C.2 Find the final internal energy

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Express the change of internal energy in terms of , , and/or .

ANSWER:

=

Correct

If you used the hints to solve this part, they lead you on a long, but instructive, path to calculate

If you used the hints to solve this part, they lead you on a long, but instructive, path to calculate . A much simpler method you might have used is to apply the first law of thermodynamics,

,

together with the fact that for an adiabatic process by definition.

Basically, the molecules transfer some of their energy and thus also momentum to the walls ofthe container, causing the expansion. So the temperature decreases, while the volume increases.Of course, some external force will eventually stop the expansion.

Internal-Combustion Engine Prototypes Ranking Task

Six new prototypes for internal-combustion engines are tested in the laboratory. For each engine, theheat energy input and output per cycle, and the designed number of cycles per second are measured.

Part A

Rank these engines on the basis of the work they perform per cycle.

Hint A.1 How to approach the problem

Engine cycles can be thought of as closed thermodynamic cycles, in which the initial and final statesof the system are the same. Thus the change in the internal energy during the cycle is zero,allowing you to directly relate the work done to the net heat transfer by applying the first law ofthermodynamics to the cycle.

Hint A.2 First law of thermodynamics applied to an engine cycle

All closed cycles involve no change in internal energy . Therefore, for net heat transfer

and net work done ,

becomes.

Based on the net heat transfer in each cycle, you should be able to correctly determine the workdone.

Hint A.3 Net heat transfer

During a simplified engine cycle, the heat transferred to the gas from the hot reservoir ( ) is

positive and the heat transferred out of the gas to the cold reservoir ( ) is negative. Thus the net

heat transfer is given by

.

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Part B

Rank these engines on the basis of their designed power output.

Hint B.1 Calculating power

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Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Part C

Rank these engines on the basis of their thermal efficiency.

Hint C.1 Definition of thermal efficiency

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Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

From Hot to Cool: A Change in Entropy

In a well-insulated calorimeter, 1.0 of water at 20 is mixed with 1.0 of ice at 0 .

Part A

What is the net change in entropy of the system from the time of mixing until the moment the

ice completely melts? The heat of fusion of ice is .

Note that since the amount of ice is relatively small, the temperature of the water remains nearlyconstant throughout the process. Note also that the ice starts out at the melting point, and you areasked about the change in entropy by the time it just melts. In other words, you can assume that thetemperature of the "ice water" remains constant as well.

Hint A.1 How to approach the problem

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Hint A.2 Description of entropy

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Hint A.3 Heat needed to melt the ice

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Express your answer numerically in joules per kelvin. Use two significant figures in youranswer.

ANSWER: = 8.35×10−2

Correct

As you would expect, in this spontaneous process the net change in entropy is positive: Theentropy increases. This is evident not just from the calculation but also from the fact that a crystalbecomes liquid and hence the degree of disorder increases.

An Expanding Monatomic Gas

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 120 . The gas

expands and, in the process, absorbs an amount of heat equal to 1200 and does an amount of work

equal to 2080 .

Part A

What is the final temperature of the gas?

Hint A.1 First law of thermodynamics

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Hint A.2 Find the change in internal energy

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Hint A.3 Calculate the change in temperature

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Use = 8.3145 for the ideal gas constant.

ANSWER:

ANSWER: = Answer not displayed

Various Gas Expansions: pV Plots and Work

An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work onthe outside world, and heat can be added or removed as necessary. The figure shows various paths thatthe gas might take in expanding from an initialstate whose pressure, volume, and temperatureare , , and respectively. The gas

expands to a state with final volume . For

some answers it will be convenient to generalizeyour results by using the variable

, which is the ratio of final to

initial volumes (equal to 4 for the expansionsshown in the figure.)The figure shows several possible paths of thesystem in the pV plane. Although there are aninfinite number of paths possible, several ofthose shown are special because one of theirstate variables remains constant during theexpansion. These have the following names:Adiabiatic: No heat is added or removed during the expansion.Isobaric: The pressure remains constant during the expansion.Isothermal: The temperature remains constant during the expansion.

Part A

Which of the curves in the figure represents an isobaric process?

ANSWER:A

B

C

D

Answer not displayed

Part B

What happens to the temperature of the gas during an isobaric expansion?

Hint B.1 Implications of

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ANSWER:Temperature increases.

Temperature remains constant.

Temperature decreases.

Answer not displayed

Part C

Which of the curves in the figure represents an isothermal process?

Hint C.1 Ideal gas law again

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ANSWER:A

B

C

D

Answer not displayed

Part D

Graphically, the work along any path in the pV plot ____________.

ANSWER:is the area to the left of the curve from to

is the area under the curve from to

requires knowledge of the temperature

Answer not displayed

Part E

Calculate , the work done by the gas as it expands along path A from to .

Hint E.1 Expression for

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Hint E.2 Find an expression for

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Hint E.3 Do the integral

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Express in terms of , , and .

ANSWER: = Answer not displayed

Part F

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Part G

Part G

Which of the curves shown represents an adiabatic expansion?

Hint G.1 Pressure and volume in an adiabatic expansion

Hint not displayed

ANSWER:A

B

C

D

Answer not displayed

A Law for Scuba Divers

SCUBA is an acronym for self-contained underwater breathing apparatus. Scuba diving has become anincreasingly popular sport, but it requires training and certification owing to its many dangers. In thisproblem you will explore the biophysics that underlies the two main conditions that may result from divingin an incorrect or unsafe manner.While underwater, a scuba diver must breathe compressed air to compensate for the increasedunderwater pressure. There are a couple of reasons for this:If the air were not at the same pressure as the water, the pipe carrying the air might close off or collapseunder the external water pressure.Compressed air is also more concentrated than the air we normally breathe, so the diver can afford tobreathe more shallowly and less often.A mechanical device called a regulator dispenses air at the proper (higher than atmospheric) pressure sothat the diver can inhale.

Part A

Suppose Gabor, a scuba diver, is at a depth of . Assume that:

The air pressure in his air tract is the same as the net water pressure at this depth. This preventswater from coming in through his nose.The temperature of the air is constant (body temperature).The air acts as an ideal gas.Salt water has an average density of around 1.03 , which translates to an increase in pressure

of 1.00 for every 10.0 of depth below the surface. Therefore, for example, at 10.0 , the net

pressure is 2.00 .

What is the ratio of the molar concentration of gases in Gabor's lungs at the depth of 15 meters tothat at the surface?

The molar concentration refers to , i.e., the number of moles per unit volume. So you are asked to

calculate .

Hint A.1 Find an equation for calculating concentrations

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Hint A.2 Find the pressure underwater

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Express your answer numerically to three significant digits.

ANSWER: = Answer not displayed

Part B

If the temperature of air in Gabor's lungs is 37 (98.6 ), and the volume is , how many moles

of air must be released by the time he reaches the surface? Let the molar gas constant be given by

= 8.31 .

Hint B.1 How to approach the problem

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Hint B.2 Solve the ideal gas law for moles

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Express your answer in moles to three significant figures.

ANSWER: = Answer not displayed

Part C

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Part D

What type of expansion does the air in the "freediver's" (no gear) lungs undergo as the diver ascends?

Hint D.1 How to approach the problem

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Hint D.2 Determine the proper prefix

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Hint D.3 Possible terms for description of processes

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Give the answer as one word.

ANSWER: Answer not displayed

An Air Conditioner: Refrigerator or Heat Pump?

The typical operation cycle of a common refrigerator is shown schematically in the figure . Both the

The typical operation cycle of a common refrigerator is shown schematically in the figure . Both thecondenser coils to the left and the evaporatorcoils to the right contain a fluid (the workingsubstance) called refrigerant, which is typically invapor-liquid phase equilibrium. The compressortakes in low-pressure, low-temperature vaporand compresses it adiabatically to high-pressure,high-temperature vapor, which then reaches thecondenser. Here the refrigerant is at a highertemperature than that of the air surrounding thecondenser coils and it releases heat byundergoing a phase change. The refrigerantleaves the condenser coils as a high-pressure,high-temperature liquid and expandsadiabatically at a controlled rate in theexpansion valve. As the fluid expands, it coolsdown. Thus, when it enters the evaporator coils,the refrigerant is at a lower temperature than itssurroundings and it absorbs heat. The air surrounding the evaporator cools down and most of therefrigerant in the evaporator coils vaporizes. It then reaches the compressor as a low-pressure, low-temperature vapor and a new cycle begins.

Part A

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has7.00 of refrigerant flowing through its circuit each cycle. The refrigerant enters the evaporator coils

in phase equilibrium, with 54.0 of its mass as liquid and the rest as vapor. It flows through the

evaporator at a constant pressure and when it reaches the compressor 95 of its mass is vapor. In

each cycle, how much heat is absorbed by the refrigerant while it is in the evaporator? The heat of

vaporization of the refrigerant is 1.50×105 .

Hint A.1 How to approach the problem

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Hint A.2 Find the percentage of refrigerant transformed to vapor

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Express your answer numerically in joules.

ANSWER: = Answer not displayed

Part B

In each cycle, the change in internal energy of the refrigerant when it leaves the compresser is1.20×105 . What is the work done by the motor of the compressor?

Hint B.1 Adiabatic compression

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Express your answer in joules.

ANSWER: = Answer not displayed

ANSWER: = Answer not displayed

Part C

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Melting Ice with a Carnot Engine

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a coldreservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heatrejected by the engine melts a mass of ice equal to 2.40×10−2 .

Throughout this problem use for the heat of fusion for water.

Part A

During this time, how much work is performed by the engine?

Hint A.1 How to approach the problem

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Hint A.2 Temperature conversion

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Hint A.3 Calculate the heat rejected

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Hint A.4 Calculate the heat absorbed

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Hint A.5 Using the first law of thermodynamics

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ANSWER: = Answer not displayed

Irreversible versus Reversible Processes

Part A

Which of the following conditions should be met to make a process perfectly reversible?

Hint A.1 Reversible processes

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Check all that apply.

ANSWER:Any mechanical interactions taking place in the processshould be frictionless.

Any thermal interactions taking place in the process shouldoccur across infinitesimal temperature or pressuregradients.

The system should not be close to equilibrium.

Answernotdisplayed

Part B

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Entropy Change of an Expanding Gas

Two moles of an ideal gas undergo a reversible isothermal expansion from 2.27×10−2 to 4.62×10−2

at a temperature of 29.5 .

Part A

What is the change in entropy of the gas?

Hint A.1 How to approach the problem

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Hint A.2 Calculate the work done by the gas

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Hint A.3 Calculating the change in entropy

Hint not displayed

Express your answer numerically in joules per kelvin.

ANSWER: = Answer not displayed

Does Entropy Really Always Increase?

An aluminum bar of mass 2.00 at 300 is thrown into a lake. The temperature of the water in the

lake is 15.0 ; the specific heat capacity of aluminum is 900 .

Part A

The bar eventually reaches thermal equilibrium with the lake. What is the entropy change of

the lake? Assume that the lake is so large that its temperature remains virtually constant.

Hint A.1 How to approach the problem

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Hint A.2 Find the heat absorbed by the lake

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Hint A.3 Entropy change in an isothermal process

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Express your answer numerically in joules per kelvin

ANSWER: = Answer not displayed

Part B

Has the entropy of the aluminum bar decreased or increased?

Hint B.1 How to approach the question

Hint not displayed

ANSWER:Since the entropy change of a system is always positive, wecan deduce that the entropy of the aluminum bar hasincreased.

Since the final lower temperature of the bar means loweraverage speed of molecular motion, we can deduce that theentropy of the bar has decreased.

We don't have enough information to determine whether theentropy of the aluminum bar has decreased or increased.

Answernotdisplayed

Part C

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Part D

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Entropy Change in a Free Expansion: A Microscopic View

A thin partition divides a thermally insulated vessel into a lower compartment of volume and an upper

compartment of volume . The lower compartment contains moles of an ideal gas; the upper part is

evacuated.

Part A

When the partition is removed, the gas expands and fills both compartments. How many moles of

gas were initially contained in the lower compartment if the entropy change of the gas in this free-expansion process is 17.28 ?

expansion process is 17.28 ?

Hint A.1 How to approach the problem

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Hint A.2 Find the number of possible microscopic states of a gas after a free expansion

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Hint A.3 Find the microscopic expression for the change in entropy of a gas

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Hint A.4 Relating to the gas constant

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Express your answer to three significant figures.

ANSWER: = Answer not displayed

Score Summary:

Your score on this assignment is 100%.You received 40 out of a possible total of 40 points.