# Ch 14.2 Help

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## Ch 14.2 Help

1) A gas does 359.4 J of work at constant temperature ("isothermal"). How much heat is exchanged with the environment (see sheet 8,12') ? Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

2) An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 1.575 atm and volume VB is 366.3 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

3) A heat engine withdraws heat at the rate of 22.72 W from a heat reservoir. What is the power output of the engine if 11.044 W of heat are released into the heat sink (see sheet 16,21) ? Indicate with a positive (negative) sign whether the input heat is always smaller (larger) than the output heat.

4) A heat engine converts 58.81 % of the input heat into mechanical work. How much heat is released into the environment when the engine does 19.37 kJ of work (see sheet 16,17) ? Indicate with a negative (positive) sign whether the the efficiency of the engine is always (not always) less than 1.

__Answer__: 359.4__How to solve__:2) An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 1.575 atm and volume VB is 366.3 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

__Answer__:__How to solve__:3) A heat engine withdraws heat at the rate of 22.72 W from a heat reservoir. What is the power output of the engine if 11.044 W of heat are released into the heat sink (see sheet 16,21) ? Indicate with a positive (negative) sign whether the input heat is always smaller (larger) than the output heat.

__Answer__: -11.675999999999998__How to solve__:4) A heat engine converts 58.81 % of the input heat into mechanical work. How much heat is released into the environment when the engine does 19.37 kJ of work (see sheet 16,17) ? Indicate with a negative (positive) sign whether the the efficiency of the engine is always (not always) less than 1.

__Answer__: -13566.575412344839__How to solve__:Last edited by periwinkle on Mon Dec 08, 2008 2:47 am; edited 1 time in total

**periwinkle**- Posts : 22

Join date : 2008-09-17

## Question 1

Question 1

The answer is obvious from the post above but I'll explain why,

According to sheet 12 of the notes, an ideal gas expands at a constant temperature, and Delta U:T = constant..so if Delta T = 0, then DeltaU = 0...

So, the amount of work done is equal to the amount of heat exchanged

The answer is obvious from the post above but I'll explain why,

According to sheet 12 of the notes, an ideal gas expands at a constant temperature, and Delta U:T = constant..so if Delta T = 0, then DeltaU = 0...

So, the amount of work done is equal to the amount of heat exchanged

**Kathleen**- Guest

## Question 3

A heat engine withdraws heat at the rate of 22.72 W from a heat reservoir. What is the power output of the engine if 11.044 W of heat are released into the heat sink (see sheet 16,21) ? Indicate with a positive (negative) sign whether the input heat is always smaller (larger) than the output heat.

Answer: -11.675999999999998

So I don't know if this is right but I used the formula from sheet 16 which is W = Qh-QL

With Qh being the input heat and QL being the output heat released into the heat sink

And I basically just did 22.72-11.044 = 11.676

And the answer is negative

Edit: I just did that with my question on maple TA and it was marked correct so that should work

Answer: -11.675999999999998

So I don't know if this is right but I used the formula from sheet 16 which is W = Qh-QL

With Qh being the input heat and QL being the output heat released into the heat sink

And I basically just did 22.72-11.044 = 11.676

And the answer is negative

Edit: I just did that with my question on maple TA and it was marked correct so that should work

**Kathleen**- Guest

## Re: Ch 14.2 Help

I can't really figure out

For

(1.504*101300)*(372.3/1000000) = 56.72184096

146.2 - 56.72184096 = 89.47815904

89.47815904/4.187 =

Any ideas? I haven't really worked on

*Question 2*(or*Question 4*)For

*Question 2*I'm not sure if what I'm doing is anywhere near correct, here's what I've done so far:(1.504*101300)*(372.3/1000000) = 56.72184096

146.2 - 56.72184096 = 89.47815904

89.47815904/4.187 =

**21.37047027**Any ideas? I haven't really worked on

*Question 4*yet.**Guest01**- Posts : 133

Join date : 2008-09-19

## Question 4

4) A heat engine converts 58.81 % of the input heat into mechanical work. How much heat is released into the environment when the engine does 19.37 kJ of work (see sheet 16,17) ? Indicate with a negative (positive) sign whether the the efficiency of the engine is always (not always) less than 1.

Answer: -13566.575412344839

From Sheet 17:

e = W/Qh

e = efficiency = 58.81/100 = .5881

W = work = 19.37kj*1000 = 19370 J

Qh = input heat

.5881 = 19370/Qh

Solve and you find Qh = 32936.57541

Then do W = Qh - QL

Where QL is the output heat

19370 J = 32926.57541 - Qh

Qh = 32926.57541 - 19370

Qh = 13566.57541

Answer is negative because the efficiency which is usually a percent is less than 1 when put back into decimal form

Answer: -13566.575412344839

From Sheet 17:

e = W/Qh

e = efficiency = 58.81/100 = .5881

W = work = 19.37kj*1000 = 19370 J

Qh = input heat

.5881 = 19370/Qh

Solve and you find Qh = 32936.57541

Then do W = Qh - QL

Where QL is the output heat

19370 J = 32926.57541 - Qh

Qh = 32926.57541 - 19370

Qh = 13566.57541

Answer is negative because the efficiency which is usually a percent is less than 1 when put back into decimal form

**Kathleen**- Guest

## Question 2

I can't seem to get question 2 either yet but this is the closest ive gotten

So, I figured you would first have to determine the amount of work done from A->B and then B->C and then add it together.

Using the info, and after converting to SI units you could do

For A->B

W=P*Change in Volume

W=152355.2*(2.722e-4) <- i found that by doing the volume of B - volume of C (100cm^3)

W= 41.4756/4.187 = 9.905

And then it says the work for B->C = 146.2/4.187 = 34.917

and then...if you subtracted the two, which I don't know why you would but if you did you got something around 25. So maybe the problem deals with calculating the work done from each point to point and adding it together?

So, I figured you would first have to determine the amount of work done from A->B and then B->C and then add it together.

Using the info, and after converting to SI units you could do

For A->B

W=P*Change in Volume

W=152355.2*(2.722e-4) <- i found that by doing the volume of B - volume of C (100cm^3)

W= 41.4756/4.187 = 9.905

And then it says the work for B->C = 146.2/4.187 = 34.917

and then...if you subtracted the two, which I don't know why you would but if you did you got something around 25. So maybe the problem deals with calculating the work done from each point to point and adding it together?

**Kathleen**- Guest

## Re: Ch 14.2 Help

How do you know the volume for C is 100 cm

Well the energy given is a combined energy, so I'm assuming whatever is missing we need to compensate for, I'm guessing that's why there's an addition to that? But wasn't that compensated for previously? Shouldn't we be subtracting then?

^{3}?Well the energy given is a combined energy, so I'm assuming whatever is missing we need to compensate for, I'm guessing that's why there's an addition to that? But wasn't that compensated for previously? Shouldn't we be subtracting then?

**Guest01**- Posts : 133

Join date : 2008-09-19

## Re: Ch 14.2 Help

if you look at the actual picture of it on maple Ta, on the graph it says volume is on x axis and it says 100 where the C is so i assumed that meant it was 100cm^3, so the only change from C->A would be in pressure

It says in the hint to account for the changes from A->b, B->C, and C->A, and you can find the change from A->B, its given for B->c, and I don't know what it would be for C->a because the volume change is 0, so multiplying any pressure by that is 0, which would make work zero (in the sense that W=p*change in volume and if W=P*0 then W=0) but that doesn't seem to quite work. From my other method my value ended up being about 25.01 which was sort of like 25.009 or whatever the value was rounded up but when i tried that method on maple ta for my own numbers it didnt quite work out

It says in the hint to account for the changes from A->b, B->C, and C->A, and you can find the change from A->B, its given for B->c, and I don't know what it would be for C->a because the volume change is 0, so multiplying any pressure by that is 0, which would make work zero (in the sense that W=p*change in volume and if W=P*0 then W=0) but that doesn't seem to quite work. From my other method my value ended up being about 25.01 which was sort of like 25.009 or whatever the value was rounded up but when i tried that method on maple ta for my own numbers it didnt quite work out

**Kathleen**- Guest

## Question 2

I am having the same problem with question number 2 I can't figure it out. If someone knows how to do it please help I am sure all of us would really appreciate it. Thank you.

**sedwards**- Guest

## Question 2

I also need help with number 2 if anyone could help I would really appreciate it I just cant figure it out.

**Jones**- Guest

## question 2 answer

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.492 atm and volume VB is 268.5 cm3. How many calories are exchanged with the environment if 140.7 J of work is done on the gas during the trip B-C (see sheet 7,? (Hint: The heat exchange Q asked for is for the complete "trip" from A to B to C to A. What does the gas equation tell you about the temperature change, and hence about the change in internal energy, for the complete trip, when the gas returns to the same PV-pair in point A (see sheet 3,5))? When calculating the work you must split the complete trip into 3 pieces, WAB+ WBC+WCA.)

Wtotal=WAB+WBC+WCA

WAB=P*deltaV

P=2.492atm*101300=252439.6Pa

At A, V=100cm^3=.000100m^3

At B, V=268.5cm^3=.0002685m^3

so deltaV=.000100-.0002685=-1.685E-4

WAB=(252439.6)(-1.685E-4)=-42.5360726J <--leave this as is, don't convert to calories just yet

WBC is given as 140.7J

WCA=0J because from point C to point A there is no change in volume, it's constant at 100cm^3

and W=P*deltaV so W=P*0=0

Wtotal=-42.5360726J + 140.7J +0J=98.1639274J

convert to calories:98.1639274J/4.187=

Wtotal=WAB+WBC+WCA

WAB=P*deltaV

P=2.492atm*101300=252439.6Pa

At A, V=100cm^3=.000100m^3

At B, V=268.5cm^3=.0002685m^3

so deltaV=.000100-.0002685=-1.685E-4

WAB=(252439.6)(-1.685E-4)=-42.5360726J <--leave this as is, don't convert to calories just yet

WBC is given as 140.7J

WCA=0J because from point C to point A there is no change in volume, it's constant at 100cm^3

and W=P*deltaV so W=P*0=0

Wtotal=-42.5360726J + 140.7J +0J=98.1639274J

convert to calories:98.1639274J/4.187=

**23.44493131calories****ryan240**- Guest

## Question 2

Just a note, The answer for Question 2 is POSITIVE, because on Maple TA it doesnt ask to indicate with a negative/positive sign or anything. Just letting you know..

**Kathleen**- Guest

## question 2

My question 2 is different from the one posted, this is how it reads....

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

how do you solve for this question?

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

how do you solve for this question?

**helpp**- Guest

## Different Question 2

I have the same question 2 that you do and I do not know how to figure it out does anyone have any ideas?

**hwilson**- Guest

## Re: Ch 14.2 Help

That's weird that you guys have a different question 2.

I think what you have to do is the same concept as posted above, because since W=p*change in V, even though they now give you a pressure for C, which changes from C to A, the volume still does not change (so I believe, I'm not sure if your picture looks different also). This means that the W will still equal 0. Try using the above formula and ignoring the additional information about C. If this doesnt work then I'll try thinking of something else.

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

I think you would still do 2.786*101300 = __

233.5 / 1000000 = 1e-4(volume at point A)-2.335e-4 = -1.335e-4

W(AB)= (2.786*101300)*(-1.335e-4) = -#

And then find the Work from B->C

Volume change would be (2.335e-4-1e-4) = 1.335e-4

And then I don't know if you would do a pressure change (as in PB-PC) or just do the 1atm*101300

but it would be Work(BC) = Pressure * 1.335e-4 = ___

and then add those two answers together and divide by 4.187 to put into calories

I don't have a calculator on me so I can't work out the problem right now but I think that should work

I think what you have to do is the same concept as posted above, because since W=p*change in V, even though they now give you a pressure for C, which changes from C to A, the volume still does not change (so I believe, I'm not sure if your picture looks different also). This means that the W will still equal 0. Try using the above formula and ignoring the additional information about C. If this doesnt work then I'll try thinking of something else.

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

I think you would still do 2.786*101300 = __

233.5 / 1000000 = 1e-4(volume at point A)-2.335e-4 = -1.335e-4

W(AB)= (2.786*101300)*(-1.335e-4) = -#

And then find the Work from B->C

Volume change would be (2.335e-4-1e-4) = 1.335e-4

And then I don't know if you would do a pressure change (as in PB-PC) or just do the 1atm*101300

but it would be Work(BC) = Pressure * 1.335e-4 = ___

and then add those two answers together and divide by 4.187 to put into calories

I don't have a calculator on me so I can't work out the problem right now but I think that should work

**Kathleen**- Guest

## Re: Ch 14.2 Help

who dont love jimbelain!!! jimbalain has all the answers nd has the biggest ....

**yeaaaaaa**- Guest

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