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Find the work and heat transferred in a carnot cycle.

  • Thread starter Kelsi_Jade
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I'm having some difficulty understanding how to find the work and heat transferred in a carnot cycle.

Problem:
One mole of ideal gas (cV=1.5 R) perform a Carnot cycle between the temperature 400 K and 300 K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume is 5 liter. Find the work performed during a cycle and the amount of heat exchanged with the two sources.

I know that since the process is isothermal the temperature remains constant: ΔT=0 but Q≠0.
Vi=1L
Vf=5L
Th=400k
Tc=300k
Qh=Th(Sb-Sa) - where Sb is the maximum system entropy and Sa is the minimum system entropy
Qc=Tc(Sb-Sa)
W=[itex]\int[/itex]PdV=(Th-Tc)(Sb-Sa)

My confusion is where I need to find the entropy to solve for the heat exchanged bt the sources is Q is needed to find entropy??

If someone could let me know if I'm going in the right direction equation-wise, or help explain where to go next it would be appreciated.
 

Answers and Replies

  • #2
TSny
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Hello, Kelsi_Jade. Welcome to PF!

For the isothermal expansion at 400K, you should be able to deduce values of the change in internal energy, ΔU, and and the work done by the gas, W. Then, apply the first law to this expansion.
 
  • #3
Andrew Mason
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I'm having some difficulty understanding how to find the work and heat transferred in a carnot cycle.

Problem:
One mole of ideal gas (cV=1.5 R) perform a Carnot cycle between the temperature 400 K and 300 K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume is 5 liter. Find the work performed during a cycle and the amount of heat exchanged with the two sources.

I know that since the process is isothermal the temperature remains constant: ΔT=0 but Q≠0.
Vi=1L
Vf=5L
Th=400k
Tc=300k
Qh=Th(Sb-Sa) - where Sb is the maximum system entropy and Sa is the minimum system entropy
Qc=Tc(Sb-Sa)
W=[itex]\int[/itex]PdV=(Th-Tc)(Sb-Sa)

My confusion is where I need to find the entropy to solve for the heat exchanged bt the sources is Q is needed to find entropy??
You just need to find Qh and Qc.

Can you express Qh in terms of Vf and Vi? (hint: how is dQ related to PdV? hint2: how is P related to V during the isothermal process?).

Once you know Qh you can find Qc from Th and Tc since ΔScycle = 0

AM
 
  • #4
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For the isothermal expansion at 400K, you should be able to deduce values of the change in internal energy, ΔU, and and the work done by the gas, W.
For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)

You just need to find Qh and Qc.
Can you express Qh in terms of Vf and Vi? (hint: how is dQ related to PdV? hint2: how is P related to V during the isothermal process?).
Once you know Qh you can find Qc from Th and Tc since ΔScycle = 0
Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?
 
  • #5
TSny
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For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)
Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?
Yes. That's right. So, you have Qh. Can you find the efficiency of the Carnot cycle from the two temperatures? If so, then you should be able to use the efficiency to find the work for the entire cycle and also QC.
 
  • #6
Andrew Mason
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For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)



Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?
Yes. Write out the expression for ΔS for the isothermal expansion and for the other three parts of the cycle. What do all those ΔS terms add up to? That will enable you to find Qc.

AM
 
  • #7
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So, you have Qh. Can you find the efficiency of the Carnot cycle from the two temperatures? If so, then you should be able to use the efficiency to find the work for the entire cycle and also QC.
By using Qh=nRTh*ln(Vf/Vi) , Qh=5,349.77J?

If efficiency=W/Qh and...
e=(Th-Tc)/Th

Then,

e=W/Qh=(Th-Tc)/Th

W=[(Th-Tc)/Th]Qh
W=(5,349.77J) [(400K-300K/400K)]=1,337.44J

Yes. Write out the expression for ΔS for the isothermal expansion and for the other three parts of the cycle. What do all those ΔS terms add up to? That will enable you to find Qc.
ΔScycle=0
ΔS=Qh/Th + Qc/Tc = 0?
So ,

Qc=(-Qh/Th)Tc

Qc=(-5,349.77J/400K)*300K= -4,012.33J
It makes sense for Qc to be (-) because it is heat that is "leaving" the system and going into the reservoir!
Is it expected that the values be so high, though?
 
Last edited:
  • #8
Andrew Mason
Science Advisor
Homework Helper
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By using Qh=nRTh*ln(Vf/Vi) , Qh=5,349.77J?

If efficiency=W/Qh and...
e=(Th-Tc)/Th

Then,

e=W/Qh=(Th-Tc)/Th

W=[(Th-Tc)/Th]Qh
W=(5,349.77J) [(400K-300K/400K)]=1,337.44J



ΔScycle=0
ΔS=Qh/Th + Qc/Tc = 0?
So ,

Qc=(-Qh/Th)Tc

Qc=(-5,349.77J/400K)*300K= -4,012.33J
It makes sense for Qc to be (-) because it is heat that is "leaving" the system and going into the reservoir!
Is it expected that the values be so high, though?
Looks right. At standard temperature and pressure, one mole will occupy 22.4 l. So this is a fairly compressed gas - high pressure. So it will do a lot of work in expanding 5 times. The ability to do this work comes from the heat flow.

AM
 

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