Mastering Physics HW 3 Ch 17, First Law of Thermodynamics

HW 3 Ch 17, First Law of ThermodynamicsDue: 11:59pm on Wednesday, September 23, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

Fast vs. Slow Tire Pumping

Imagine the following design for a simple tire pump. The pump is filled with a volume of air at atmospheric pressure and ambient temperature . When you push the pump handle, the air is compressed to a new (smaller) volume , raising its pressure. A valve is then opened, allowing air to flow from the pump into the tire until the remaining air in

the pump reaches the pressure of the air in the tire, .

In this problem, you will consider whether you can get more air into the tire per pump cycle by pushing the pump handlequickly or slowly.

We will make the following simplifying assumptions:

The pressure of the air in the tire, , does not change significantly as air flows into the tire from thepump.The temperature of the air in the pump does not change significantly while air is flowing into the tire (i.e.,this is an isothermal process).The air is mainly composed of diatomic molecules with .

Part A

First imagine that you push the pump handle quickly, so that the compression of air in the pump occursadiabatically. Find the absolute temperature of the air inside the pump after a rapid compression from volume to volume , assuming an ambient temperature of .

Express your answer in terms of , , , and

Hint 1. Properties of an adiabatic process

For an adiabatic process the product is constant. Using the ideal gas law, you can derive theequivalent condition that the product is constant throughout the adiabatic process.

ANSWER:

Correct

Part B

Once the tire and pump pressures have equilibrated at , what fraction of the gas particles initially in the pumpwill have ended up in the tire?

Express the fraction in terms of , , , , and . The temperatures and should not appear inyour answer.

Vipa TaVf

pt

pt

γ = 1.40

TfVi Vf Ta

Ta Vi Vf γ

pV γ

TV γ−1

= Tf Ta( )Vi

Vf

γ−1

pt f

pt pa Vf Vi γ Ta Tf

Hint 1. How to approach the problem

Find an expression for the number of particles in the pump initially ( ) and another for the number in thepump after the pump and tire have come into pressure equilibrium ( ). Use these to determine thefraction of the particles in the pump that are transferred into the tire:

.

Hint 2. Find an expression for the number of particles in a gas

Using the ideal gas law, find an expression for the number of particles in a gas at pressure , volume ,and temperature .

Use for Boltzmann's constant.

ANSWER:

Hint 3. Find

What is the ratio of the number of particles in the pump after it comes into pressure equilibrium with the tireto the number of gas particles initially in the pump (before pushing the handle)?

Express the ratio in terms of , , , , and . The temperatures and should not appearin your answer.

Hint 1. Find the initial number of particles in the pump

How many gas particles were in the pump initially?

Express your answer in terms of quantities given in the problem introduction andBoltzmann's constant .

ANSWER:

Hint 2. Find the final number of particles in the pump

How many gas particles are in the pump after the pump and tire have equilibrated?

Express your answer in terms of quantities given in the problem introduction andBoltzmann's constant . should not appear in your answer.

Hint 1. Some helpful quantities

Recall that the final pressure is and the temperature continues to be .

ANSWER:

NiNf

f = = 1−−Ni NfNi

NfNi

N p VT

kB

= N

/Nf Ni


Ni

kB

= Ni

Nf

kB Tf

pt Ta( )Vi

Vf

γ−1

ANSWER:

ANSWER:

Correct

Part C

Now imagine that you push the pump handle slowly, so that the compression of air in the pump occursisothermally. Find the absolute temperature of the air inside the pump assuming an ambient temperature of .

Hint 1. Properties of an isothermal process

Isothermal means "at constant temperature."

ANSWER:

Correct

Part D

Once the tire and pump pressures have equilibrated at , what fraction of the gas particles initially in the pumpwill have ended up in the tire?

Express the fraction in terms of , , , , and . The temperatures and should not appear inyour answer.


Find an expression for the number of particles in the pump initially ( ) and another for the number in thepump after the pump and tire have come into pressure equilibrium ( ). Use these to determine thefraction of the particles in the pump that are transferred into the tire:

= Nf

= Nf

Ni

= f 1− pt

pa( )Vf

Vi

γ

Tf Ta

= Tf Ta

pt f


NiNf

−N N N

.

Hint 2. Find an expression for the number of particles in a gas

Using the ideal gas law, find an expression for the number of particles in a gas at pressure , volume ,and temperature .

Use for Boltzmann's constant.

ANSWER:

Hint 3. Find

What is the ratio of the number of particles in the pump after it comes into pressure equilibrium with the tireto the number of gas particles initially in the pump (before pushing the handle)?

Express the ratio in terms of , , , , and . The temperatures and should not appearin your answer.

Hint 1. Find the initial number of particles in the pump

How many gas particles were in the pump initially?

Express your answer in terms of quantities given in the problem introduction andBoltzmann's constant .

ANSWER:

Hint 2. Find the final number of particles in the pump

How many gas particles are in the pump after the pump and tire have equilibrated?

Express your answer in terms of quantities given in the problem introduction andBoltzmann's constant . should not appear in your answer.

ANSWER:

ANSWER:

ANSWER:

f = = 1−−Ni Nf

Ni

Nf

Ni

N p VT

kB

= N

/Nf Ni


Ni

kB

= Ni

Nf

kB Tf

= Nf

= NfNi

Correct

Part E

Assume that and . Which method, fast pumping (adiabatic process) or slow pumping(isothermal process) will put a larger amount of air into the tire per pump cycle?

Hint 1. Adiabatic process

What is the numerical value of for the fast (adiabatic) process?

Express your answer numerically, to two significant figures.

ANSWER:

Hint 2. Isothermal process

What is the numerical value of for the slow (isothermal) process?

Express your answer numerically, to two significant figures.

ANSWER:

ANSWER:

CorrectSo when you pump quickly, not only do you get more pump cycles per unit time, but you also put more air inthe tire per cycle (at least according to this simplified model). Of course you have to pump not only faster butalso harder, since the average pressure you will pump against will be higher. In fact this higher pressure is themain reason that more particles are transferred to the tire each time.

PSS 17.1: Work in Idealgas Processes

Learning Goal:

= f 1− pt

pa

Vf

Vi

= 3pt pa = /6Vf Vi

f

= f

f

= f

fast pumping

slow pumping

To practice ProblemSolving Strategy 17.1: Work in Idealgas Processes.

A cylinder with initial volume contains a sample of a gas at pressure . On one end of the cylinder, a piston is letfree to move so that the gas slowly expands in such a way that its pressure is directly proportional to its volume. Afterthe gas reaches the volume and pressure , the piston is pushed in so that the gas is compressed isobarically toits original volume . The gas is then cooled isochorically until it returns to the original volume and pressure.Find the work done on the gas during the entire process.

MODEL Assume that the gas is ideal and the process is quasistatic.

VISUALIZE Show the process on a pV diagram. Note whether it happens to be one of the basic gas processes:isochoric, isobaric, or isothermal.

SOLVE Calculate the work as the area under the pV curve either geometrically or by carrying out the integration:

ASSESS Check your signs.

when the gas is compressed. Energy is transferred from the environment to the gas. when the gas expands. Energy is transferred from the gas to the environment.

No work is done if the volume doesn't change, .

Model

Assume the gas is ideal. Since the processes take place slowly, you can also assume they are quasistatic.

Visualize

Part A

Which of the following pV diagrams correctly represents the entire process described in this problem?

ANSWER:

V p

3V 3pV

W

work done on the gas W = − p dV∫ VfVi

= −(area under pV curve)

W > 0W < 0

W = 0

Correct

The entire process consists of three intermediate processes that take the gas from its initial state at pressure and volume and return it to the same state. As for any multistep processes that have the same initial and

final state, the pV diagram is a closed curve. Be sure to clearly identify the direction and type of each step ofthe process. You may want to mark this information on your own pV diagram.

Solve

Part B

Find the amount of work done on the gas during the entire process.

Express your answer in terms of some or all of the variables and .


p V

W

p V

Since the entire process consists of three separate steps, the total work done on the gas is given by thesum of the amounts of work done on the gas during each step of the process. For each step, you cancalculate the work as the negative of the area under the corresponding branch of the pV curve eithergeometrically or by carrying out the appropriate integration. Be careful with signs. You may want to reviewhow to distinguish between positive and negative areas.

Hint 2. Positive and negative areas

When a gas expands from an initial volume to a larger volume , the area under the curve ispositive,

When a gas is compressed from an initial volume to a smaller volume , the calculation of the areaunder the pV curve is a little trickier because it requires to integrate “backward” along the V axis. Youlearned in calculus that integrating from a larger limit to a smaller limit gives a negative result, so in thiscase the area under the pV curve is a negative area. We express this mathematically as

Hint 3. Find the work done during expansion from to

Find , the work done on the gas as it expands from volume to volume .

Express your answer in terms of and .

Hint 1. How to compute the work

Identify the branch of the pV curve that corresponds to this process. Compute the area under thatportion of the curve either geometrically or by carrying out the appropriate integration. Be careful todistinguish between positive and negative areas. Finally, the work done on the gas during thisprocess will be equal to the negative of that area. Note that in this particular problem, where thecurve of interest is a straight line, it is easier to use geometry, rather than calculus, to computeareas.

ANSWER:

Hint 4. Find the work done during isobaric compression

Find , the work done on the gas during isobaric compression.

Express your answer in terms of and .

Hint 1. How to compute the work

Identify the branch of the pV curve that corresponds to this process. Compute the area under thatportion of the curve either geometrically or by carrying out the appropriate integration. Be careful todistinguish between positive and negative areas. Finally, the work done on the gas during thisprocess will be equal to the negative of that area. Note that in this particular problem, where thecurve of interest is a straight line, it is easier to use geometry, rather than calculus, to computeareas.

W

Vi Vf

(area under pV curve) = p dV∫ VfVi

Vi Vf

(area under pV curve) = p dV = − p dV∫ VfVi

∫ Vi

Vf

V 3V

We V 3V

p V

= We

Wp

p V

ANSWER:

Hint 5. Find the work done during isochoric process

Find , the work done on the gas during isochoric process.

Express your answer in terms of and , or appropriate constants.

Hint 1. Work in isochoric processes

Recall that work is the negative integral of . Since the volume is not changing during anisochoric process, what value will the integrand have at any instant in this process? Whatwould then the integral of be equal to?

ANSWER:

ANSWER:

Correct

Assess

Part C

To assess your calculations done in Part B, identify the net energy transfer taking place in each step of theprocess.Sort each intermediate step of the process described in the problem introduction according to whether the netenergy transfer is to the environment or to the gas.

ANSWER:

= Wp

WV

p V

pdVpdV

pdV

= WV

= W 2pV

Correct

In both the initial expansion and the isobaric process, the gas interacts with the environment bothmechanically (the gas expands when the piston is allowed to move and is compressed when the piston ispushed in) and thermally (the gas gains or loses thermal energy). Since we can tell what is happening to theoverall energy of the gas by paying attention to the temperature we can tell which part (work or heating) wasthe dominant process in each part.

Isobaric, Isochoric, Isothermal, and Adiabatic Processes

Learning Goal:

To recognize various types of processes on diagrams and to understand the relationship between diagramgeometry and the quantities , , and .

The first law of thermodynamics is an expression of conservation of energy. This law states that changes in the internalenergy of a system can be explained in terms of energy transfer into or out of the system in the form of heat and/or work .

In this problem, we will write the first law of thermodynamics as

.

Here "in" means that energy is being transferred into the system by atomiclevel collisions, thereby raising its internalenergy ( ), and "out" means that energy is leaving the system, thereby reducing its internal energy ( ). "On"means that energy is being transferred into the system by forces in a mechanical interaction, thereby raising theinternal energy ( ), and "by" means that energy is leaving the system, thereby reducing the internal energy (

). You will determine the sizes of these energy transfers and classify their effect on the system as energy in orenergy out.

Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy

pV pVQ W ΔEth

ΔEth QW

Δ = W + QEth

Δ = + + +Eth Won Wby Qin Qout

Q > 0 Q < 0

W > 0W < 0

Zombie Rae

Stamp

can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy canalso be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to thesystem in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes thepiston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of workdone.The internal energy of an ideal gas is directly proportional to its absolute temperature . An ideal gas also obeys theideal gas law

,

so the absolute temperature is directly proportional to the product of the absolute pressure and the volume . Here denotes the amount of gas in moles, which is a constant because the gas is confined, and is the universal gas

constant.

A diagram is a convenient way to track the pressure andvolume of a system. Energy transfers by heat and/or workare associated with processes, which are lines or curves onthe diagram taking the system from one state (i.e., onepoint on the diagram) to another. Work correspondsgeometrically to the area under the curve on a diagram. Ifthe volume increases (i.e., the system expands) the work willbe classified as an energy output from the system.

Part A

What is the sign of as the system of ideal gas goes from point A to point B on the graph? Recall that isproportional to .


Use the ideal gas law to figure out how the absolute temperature of the gas in state A compares to itsabsolute temperature in state B. Since the internal energy of an ideal gas is proportional to its absolutetemperature, this will tell you how changes from state A to state B.

ANSWER:

T

pV = nRT

T p Vn R

pV

pV

pV

ΔEth Eth

T

Eth

The internal energy of the system increases, so is positive.

The internal energy of the system decreases, so is negative.

The states A and B have the same internal energy, so is zero.

cannot be determined without knowing the process used (i.e., the path taken) to get from state Ato state B

ΔEth

ΔEth

ΔEth

ΔEth

Correct

The value of depends only on the state of the system. Thus depends only on the endpoint states,not on the process followed that determines the path between the endpoint states.

One possible way for the system to get from state A to state B is to follow a hyperbolic curve through point C, alongwhich the product of is a constant. Temperature isproportional to the product , so this is a constanttemperature path, also known as an isothermal process.

Part B

How are and related during this isothermal expansion?

Hint 1. Find the sign of

is defined as the work done on the system, which in this case is the gas. Recall that the magnitude ofthe work done by the gas, , in going from one state of the gas to another is the area underneath thecurve defined by the path. If the work done by the gas is nonzero, then the sign of is determined bythe direction of the path. Since , which of the following describes in Part B?

ANSWER:

ANSWER:

Eth ΔEth

pVpV

Q W

W

WWgas

Wgas

W = −Wgas W

W < 0W = 0W > 0

Correct

You can tell that the system is losing internal energy due to work because its volume is increasing. Theinternal energy change during any isothermal process involving an ideal gas is zero, so here the system mustgain as much energy in the form of heat as it loses by doing work during this process.

Another way to get from state A to state B is to go vertically from A to point D, holding volume constant, and then gohorizontally to point B, holding pressure constant. A constantvolume path is called an isochoric process. A constantpressure path is called an isobaric process.

Part C

How are and related during the isochoric part of the overall path from state A to state D?


Use the ideal gas law to determine how the absolute temperature of the gas in state A compares to itsabsolute temperature in state D. This will help you determine whether the net energy transfer is in or out,since the internal energy of an ideal gas is proportional to its absolute temperature.

ANSWER:

Both and equal zero.

Both and provide energy input.

Both and provide energy output.

provides energy output, while provides energy input. They are equal in magnitude.

provides energy input, while provides energy output. They are equal in magnitude.

W Q

W Q

W Q

W Q

W Q

Q W

Correct

You can tell that the system is losing internal energy since its temperature goes down (since goes down).No work is done during any isochoric process, since no area accumulates under a vertical curve. Henceenergy transfer in the form of heat must account for the entire internal energy change.

Part D

How are and related during the isobaric part of the overall path from state D to state B?

ANSWER:

Correct

In going from state A to state D the system loses internal energy. Since the overall change of internal energyfrom state A to state B is zero, during the isobaric part of the overall process the system internal energy mustincrease. Since the system is expanding, internal energy is lost from the system due to work. Hence mustexceed (in magnitude) to explain the net increase in internal energy.

Another way to get from state A to state B is to follow an adiabatic path from state A to state E, in which no heatenergy transfer is allowed, and then to follow an isochoric path from state E vertically to state B. Notice that during theadiabatic part of this path, from state A to state E, by definition and internal energy is lost due to work since thesystem is expanding.

Both and equal zero.

provides energy input, while equals zero.

provides energy output, while equals zero.

provides energy input, while provides energy output.

provides energy output, while provides energy input.

Q W

Q W

Q W

Q W

Q W

pV

Q W

Both and provide energy input.

Both and provide energy output.

provides energy output, while provides energy input. They are equal in magnitude.

provides energy output, while provides energy input; is larger.

provides energy output, while provides energy input; is larger.

W Q

W Q

W Q

W Q W

W Q Q

QW

Q = 0

Part E

Which of the following statements are true about the isochoric part of the overall path, from state E to state B?

Check all that apply.


Recall that the total internal energy change from state A to state B is zero. This means that the isochoricprocess must undo any changes to internal energy made during the adiabatic process.

ANSWER:

CorrectSince no work is allowed in isochoric processes, must serve as an energy input to explain the increase inboth absolute temperature and internal energy.

One more way to get from state A to state B is to follow a direct path through state F. This process is not isobaric,isochoric, isothermal, or adiabatic, yet you can draw some conclusions about its energetics using the first law ofthermodynamics.

is zero.

provides energy input.

decreases.

increases.

W

Q

T

Eth

Q

Part F

Which of the following statements are true about the first half of this process, just going from state A to state F?

Check all that apply.

ANSWER:

Correct

State F has a larger value than state A, so the internal energy increases in this part of the process. Sincethe system is expanding, internal energy is lost from the system due to work. Hence must exceed (inmagnitude) to explain the net increase in internal energy.

Understanding what happens during the second half of the process, going from state F to state B, is moresubtle. The temperature and the internal energy both go down. Since the system continues to expand, provides energy output. However, it is challenging to determine whether provides energy input or energyoutput from state F to state B. Can you figure it out?

Simple Ways of Expanding

A plot of pressure as a function of volume is known as a pV diagram. pV diagrams are often used in analyzingthermodynamic processes. Consider an ideal gas that starts in state O, as indicated in the diagram. Your task is todescribe how the gas proceeds to one of four different states, along the different curves indicated.

Although there are an infinite number of such curves, several are particularly simple because one quantity or anotherdoes not change:

Adiabatic: . No heat is added or subtracted.Isothermal: . The temperature does not change. (The prefix "iso" means equal or alike.)

Both and increase.



is larger (in magnitude) than .

T Eth

W

Q

Q W

pVQ W

WQ

ΔQ = 0ΔT = 0

Δ = 0

Isobaric: . The pressure does not change. (Abarometer measures the pressure.)Isochoric: . The volume does not change.(This process is infrequently used.)

The key idea in determining which of these processes isoccurring from a pV plot is to recall that an ideal gas mustobey the ideal gas equation of state: (wherethe constant is the Boltzmann constant, which has thevalue in SI units). Generally , thenumber of gas particles, is held constant, so you candetermine what happens to at various points along thecurve on the pV diagram.

Note that in this problem, as is usually assumed, theprocesses happen slowly enough that the gas remains inequilibrium without hot spots, without propagating pressure waves from a rapid change in volume, or without involvingsimilar nonequilibrium phenomena. Indeed, the word "adiabatic" is often used by scientists to describe a process thathappens slowly and smoothly without irreversible changes in the system.

Part A

What type of process does curve OA represent?

Hint 1. What quanitity remains constant?

Curve OA is horizontal; the pressure of the system remains constant throughout this process. What is thename of a process in which pressure remains constant?

ANSWER:

Correct

Part B

What type of process does curve OC represent?

Hint 1. Relationship between pressure and volume

When , ; when , . Evidently, pressure is proportional to theinverse of the volume. What is the name of a process in which is proportional to ?

Δp = 0

ΔV = 0

pV = N TkB

kB1.381 × J/K10−23 N

T

adiabatic

isobaric

isochoric

isothermal

V / = 2V0 p/ = 1/2p0 V / = 4V0 p/ = 1/4p0p 1/V

Hint 2. Use the ideal gas law

Recall the ideal gas equation of state for a fixed amount of gas: , where is some constant.Solving this equation for yields . If is to be proportional to , the temperature mustremain constant throughout the process. What is the name of a process in which temperature remainsconstant?

ANSWER:

Correct

Detailed analysis of curve OB

The following questions refer to the process represented by curve OB, in which an ideal gas proceeds from state O tostate B.

Part C

The pressure of the system in state B is __________ the pressure of the system in state O.

ANSWER:

Correct

Part D

The work done by the system is __________.

Hint 1. How to find the work done using the pV diagram

On a pV diagram, the work done on the system during a particular process is represented by the negative ofthe area under the curve describing that process. The work done by the system would be the opposite ofthis. Is the area under curve OB greater than, less than, or equal to zero?

ANSWER:

pV = cT cp p = cT/V p 1/V T

adiabatic

isobaric

isochoric

isothermal

greater than

less than

equal to

Correct

Part E

The temperature of the system in state B is __________ the temperature of the system in state O.

Hint 1. Compare curve OB to an isothermal process

Does curve OB lie above or below a curve representing an isothermal (constanttemperature) process? Usethis, along with the ideal gas equation of state, to figure out how changes as the system proceeds fromstate O to state B.

Hint 2. Computing the change in temperature mathematically

The ideal gas equation of state for a fixed amount of gas is , where is some constant. To findthe sign of the change in temperature, you could find at point B, then subtract the value of

at point O. Although this method will not yield the actual change in temperature, it will givea number proportional to the change in temperature (with the proper sign).

ANSWER:

Correct

Part F

The internal energy of the system in state B is __________ the internal energy of the system in state O.

Hint 1. Relationship between internal energy and temperature

The change in internal energy of an ideal gas is proportional to its change in temperature; the constant ofproportionality is the heat capacity at constant volume.

ANSWER:

greater than zero

less than zero

equal to zero

T

pV = cT c(p/ )(V / )p0 V0

(p/ )(V / )p0 V0

greater than

less than

equal to

Correct

Piston in Water Bath Conceptual WorkEnergy Problem

Imagine a piston containing a sample of ideal gas in thermal equilibrium with a large water bath. Assume that the pistonhead is perfectly free to move unless locked in place, and the walls of the piston readily allow the transfer of energy viaheat unless wrapped in insulation. The piston head is unlocked and the gas is in an equilibrium state.

For each of the actions described below, state whether the work done by the gas, the heat energy transferred tothe gas, and the change in the internal energy of the gas are positive ( ), negative ( ), or zero (0).

After each action the piston is reset to its initial equilibrium state.

Part A

Action: Lock the piston head in place. Hold the piston above a very hot flame.

Enter the signs of , , and . Use , , or 0 separated by commas. For example, if is positive, is negative, and is zero, you would type +,‐,0.

Hint 1. Find the sign of the work done by the gas

With the piston head locked in place, will the work done by the gas be positive, negative, or zero?

Give the sign of . Answer with +, ‐, or 0.

Hint 1. Work done by a gas

When a gas expands, it does work on its surroundings. When a gas is compressed, its surroundingsdo work on it. Considering the gas as the system of interest, when the gas does work, this work isconsidered positive, and when the work is done on the gas, this work is considered negative.

ANSWER:

Hint 2. Find the sign of the heat transferred to the gas

With the piston held above a flame, will the heat transferred to the gas be positive, negative, or zero?


Hint 1. Heat energy transferred to a gas

greater than

less than

equal to

W QEth + −

W Q Eth + − W QEth

W

Q

Heat energy can either flow into or out of a gas sample. When the energy flows into the gas, it isconsidered positive heat; when the energy flows out of the gas it is considered negative heat.

ANSWER:

Hint 3. Find the sign of the change in internal energy of the gas

Based on the signs of work and heat , what must be the sign of change in internal energy ?


Hint 1. First law of thermodynamics

The first law of thermodynamics states that the change in internal energy of a gas is equal tothe heat added to the gas minus the work done by the gas:

.This relationship must hold for each of the actions described.

ANSWER:

ANSWER:

Correct

Part B

Action: Very slowly push the piston head down.



If the piston head is pushed down, will the work done by the gas be positive, negative, or zero?


Hint 1. Work done by a gas

When a gas expands, it does work on its surroundings. When a gas is compressed, its surroundingsdo work on it. Considering the gas as the system of interest, when the gas does work this work is

W Q Eth

Eth

EthQ W

= W + QEth

0,+,+


W

considered positive, and when the work is done on the gas this work is considered negative.

ANSWER:


Since the piston is in constant thermal equilibrium with a large water bath, will the change in internal energyof the gas be positive, negative, or zero?


Hint 1. Change in internal energy of a gas

The internal energy of a gas is directly proportional to its temperature. Therefore, if the gastemperature increases, its change in internal energy is positive. If the temperature decreases, thechange in internal energy of the gas is negative.

ANSWER:


Based on the signs of and , what must be the sign of ?


Hint 1. First law of thermodynamics

The first law of thermodynamics states that the change in internal energy of a gas is equal tothe heat added to the gas minus the work done by the gas:

.This relationship must hold for each of the actions described.

ANSWER:

ANSWER:

Correct

Eth

W Eth Q

Q

EthQ W

= W + QEth

+,,0

Part C

Action: Lock the piston head in place. Plunge the piston into very cold water.



With the piston head locked in place, will the work done by the gas be positive, negative, or zero?


ANSWER:


After the piston is plunged into cold water, will the heat transferred to the gas be positive, negative, or zero?


ANSWER:



Give the sign of ; Answer with +, ‐, or 0.

ANSWER:

ANSWER:

Correct

Part D

Action: Wrap the piston in insulation. Pull the piston head up.



W

Q

W Q Eth

Eth

0,,



If the piston head is pulled up, will the work done by the gas be positive, negative, or zero?


ANSWER:


The piston is wrapped in insulation; will the heat transferred to the gas be positive, negative, or zero?


ANSWER:




ANSWER:

ANSWER:

Correct

pV Diagram for a Piston

A container holds a sample of ideal gas in thermal equilibrium, as shown in the figure. One end of the container issealed with a piston whose head is perfectly free to move, unless it is locked in place. The walls of the containerreadily allow the transfer of energy via heat, unless the piston is insulated from its surroundings.

W

Q

W Q Eth

Eth

,0,

Refer to the pV diagram presented to answer the questions below. In each case, the piston head is initially unlockedand the gas is in equilibrium at the pressure and volumeindicated by point 0 on the diagram.

Part A

Starting from equilibrium at point 0, what point on the pV diagram will describe the ideal gas after the followingprocess?"Lock the piston head in place, and hold the container above a very hot flame."

Hint 1. Understand the graph

The graph axes represent the pressure and volume of the gas sample. If either of these quantities changes,the position of the point representing the gas sample must also change. Moreover, since temperature isproportional to the product of pressure and volume, the temperature of the gas can also be determined fromthe graph.A common way of denoting temperature on a pV graph is by including curves representing constanttemperature, curves on which the product of pressure and volume are constant. These are called isotherms(from the Greek iso, equal, and therme, heat), and three isotherms are indicated on the graph. Of the threeisotherms, which one designates the largest temperature, , , or ?

ANSWER:

T1 T2 T3

Hint 2. Find the change in volume

With the piston head locked in place, will the volume of the gas increase, decrease, or stay the same whenthe piston is placed above the flame?

ANSWER:

ANSWER:

Correct

Part B

Starting from equilibrium at point 0, what point on the pV diagram will describe the ideal gas after the followingprocess?"Immerse the container into a large water bath at the same temperature, and very slowly push the piston headfurther into the container."


If the piston head is pushed further into the container, will the volume of the gas increase, decrease, or staythe same?

ANSWER:

T1

T2

T3

The volume increases.

The volume decreases.

The volume stays the same.

point 1

point 2

point 3

point 4

point 5

point 6

point 7

point 8

Hint 2. Find the change in temperature

If the piston head is pushed into the container very slowly, and the container remains in contact with thelarge water bath, will the temperature of the gas in the container increase, decrease, or stay the same?

ANSWER:

ANSWER:

Correct

Part C

Starting from equilibrium at point 0, what point on the pV diagram will describe the ideal gas after the followingprocess?"Lock the piston head in place and plunge the container into water that is colder than the gas."

ANSWER:

The volume will increase.

The volume will decrease.

The volume will stay the same.

The temperature will increase.

The temperature will decrease.

The temperature will stay the same.

point 1

point 2

point 3

point 4

point 5

point 6

point 7

point 8

Correct

Part D

Starting from equilibrium at point 0, what point on the pV diagram will describe the ideal gas after the followingprocess?"The piston is now insulated from its surroundings. Pull the piston head further out of the container."


If the piston head is pulled further out of the container, will the volume of the gas increase, decrease, orstay the same?

ANSWER:

Hint 2. Find the change in temperature

If the piston head is pulled further out of the container, and the piston is insulated from its surroundings, willthe temperature of the gas increase, decrease, or stay the same?

ANSWER:

ANSWER:

point 1

point 2

point 3

point 4

point 5

point 6

point 7

point 8

The volume will increase.

The volume will decrease.

The volume will stay the same.

The temperature will increase.

The temperature will decrease.

The temperature will stay the same.

Correct

The First Law of Thermodynamics Derived

Learning Goal:

To understand the first law of thermodynamics and its origin.

By relating heat, thermal energy, and work, the first law lays the groundwork for thermodynamics.The first law of thermodynamics generalizes the concept of energy conservation to include heat energy. You areprobably already aware that loss of total mechanical energy (e.g., from nonconservative forces such as friction) doesnot destroy energy, but rather converts mechanical energy into thermal energy. This process, as well as the reverseprocess (conversion of thermal energy into mechanical energy), can be described quantitatively by the first law.

Like the law of mechanical energy conservation that it generalizes, the first law relates the changes in energy thatoccur from the beginning to the end of some process. The first law involves the following physical quantities:

: work done on the system by the outside world,: heat added to the system by the outside world, and

: thermal energy change of the system.

You need to look carefully at the wording used here; some other disciplines may use other definitions.

Part A

Which of the following is the sign convention that results from the above definitions?

ANSWER:

Correct

point 1

point 2

point 3

point 4

point 5

point 6

point 7

point 8

WQΔEth

is positive when the system is compressed, and is positive when heat is added to the system.

is positive when the system expands, and is positive when heat is added to the system.

is positive when the system is compressed, and is positive when heat is taken from the system.

is positive when the system expands, and is positive when heat is taken from the system.

W Q

W Q

W Q

W Q

Part B

Using your knowledge of energy conservation, express in terms of and .


The first law of thermodynamics involves heat, thermal energy, and work. Therefore, you are asked merelyto determine the correct signs for and . To help your thinking, imagine that the system is an an idealgas.

Hint 2. Determine how thermal energy and heat are related

If heat is added to a system ( ) and no work is done on the system ( ), then the thermal energy must __________.

ANSWER:

Hint 3. Determine how work and thermal energy are related

If work is done on the system ( ) and no heat is exchanged by the system ( ), then the thermalenergy must __________.

ANSWER:

ANSWER:

Correct

If this problem seemed a bit on the easy side, you're right. Essentially, all this problem asks is to expressenergy conservation using the given definitions of , , and as an equation that happens to be thefirst law of thermodynamics.

Two Closed Thermodynamic Cycles Conceptual Question

Two closed thermodynamic cycles are illustrated in the figure. The ideal gas sample can be processed clockwise or

ΔEth Q W

Q W

Q > 0 W = 0Eth

increase

decrease

remain constant

W > 0 Q = 0Eth

increase

decrease

remain constant

= ΔEth W + Q

Q W ΔEth

counterclockwise through either cycle.

Part A

Imagine processing the gas clockwise through Cycle 1. Determine whether the change in internal energy of the gasin the entire cycle is positive, negative, or zero.

Choose the correct description of for Cycle 1.

Hint 1. Closed thermodynamic cycles

A series of actions undertaken with a gas sample that result in the sample returning to its initial state form aclosed thermodynamic cycle. From steam engines to air conditioners, many machines employ closedthermodynamic cycles. Since the cycle can be endlessly repeated, the precise starting point of the cycle isnot necessary for its analysis.

Hint 2. Change in internal energy in a closed cycle

The change in internal energy of a gas depends only on the initial and final states of the gas, not on theprocess that takes the gas from one state to another. This fact allows you to determine the change ininternal energy for any closed cycle.

Hint 3. Initial and final state in a closed cycle

In any closed cycle, the initial and final states of the gas are the same. This is what allows the cycle to berepeated.

ANSWER:

Eth

positive

zero

negative

cannot be determined

Correct

Part B

Imagine processing the gas clockwise through Cycle 1. Determine whether the work done by the gas in the entirecycle is positive, negative, or zero.


Hint 1. Work in a closed cycle

Work depends on the precise path from initial state to final state. To determine the work done in a closedcycle, you must sum the work done in each portion of the cycle.

Hint 2. Compare the work done in different parts of a closed cycle

The figure shows a plot depicting a closed thermodynamic cycle. Is the magnitude of the workdone by the gas in portion 2 greater than, less than, or the same as the magnitude of the work done inportion 4 of the cycle? The parts of the cycle are labeled clockwise beginning from the left side of the cycle.

Choose the correct comparison between the value of in portion 2 and that of in portion 4 ofthe cycle.

Hint 1. Magnitude of work done

The magnitude of the work done by the gas during a portion of the cycle is equal to the area underthe curve on the pVdiagram for that portion.

ANSWER:

Wclockwise

pV W2W4

W2 W4

ANSWER:

Correct

Part C

Imagine processing the gas clockwise through Cycle 1. Determine whether the heat energy transferred to the gasin the entire cycle is positive, negative, or zero.


Hint 1. Heat transfer in a closed cycle

The easiest way to determine the heat transferred in a closed cycle is by applying the first law ofthermodynamics to the cycle.

Hint 2. Simplifying the first law of thermodynamics

All closed cycles involve no change in internal energy. Therefore,

becomes

where is the work done on the gas. Based on the work done in each cycle, you should be able tocorrectly determine the heat transfer.

ANSWER:

>

=

< W2 W4

positive

zero

negative


Qclockwise

Q = − WEth

Q = −W = WgasWgas

positive

zero

negative


Correct

Part D

Imagine processing the gas clockwise through Cycle 1 and then counterclockwise through Cycle 1. Compare thesetwo processes on the basis of the work done by the gas in the entire cycle.

Choose the correct comparison symbol.

Hint 1. Find the magnitude of the work done

How does the magnitude of the work done in going around the cycle clockwise compare to the magnitude intraversing the cycle counterclockwise?

Compare to .

ANSWER:

ANSWER:

Correct

Part E

Imagine processing the gas clockwise through Cycle 1 and then counterclockwise through Cycle 1. Compare thesetwo processes on the basis of the heat energy transferred to the gas in the entire cycle.


Hint 1. Heat transfer in a closed cycle

The easiest way to determine the heat transferred in a closed cycle is by applying the first law ofthermodynamics to the cycle. Recall that , where is the change in internal energy, is the work done on the gas ( is the work done by the gas), and is the heat transferred into the gas.

ANSWER:

Wclockwise Wcounterclockwise

>

=

< Wclockwise Wcounterclockwise

for Cycle 1

>

=

< for Cycle 1.Wclockwise Wcounterclockwise

= W + QEth Eth W−W Q

Correct

Part F

Imagine processing the gas clockwise through Cycle 1 and then clockwise through Cycle 2. Compare these twoprocesses on the basis of the work done by the gas in the entire cycle.


ANSWER:

Correct

Part G

Imagine processing the gas clockwise through Cycle 1 and then clockwise through Cycle 2. Compare these twoprocesses on the basis of the heat energy transferred to the gas in the entire cycle.


Hint 1. Find the magnitude of work done

How does the magnitude of the work done in Cycle 1 clockwise compare to the magnitude in Cycle 2clockwise?

Compare in Cycle 1 to in Cycle 2.

ANSWER:

ANSWER:

for Cycle 1

>

=

< for Cycle 1.Qclockwise Qcounterclockwise

for Cycle 1

>

=

< for Cycle 2Wclockwise Wclockwise

Wclockwise Wclockwise

in Cycle 1

>

=

< in Cycle 2.Wclockwise Wclockwise

Correct

Score Summary:Your score on this assignment is 96.5%.You received 7.72 out of a possible total of 8 points.

for Cycle 1

>

=

< for Cycle 2Qclockwise Qclockwise

Documents

Mastering Physics HW 3 Ch 17, First Law of Thermodynamics