Go Back   Physics Help Forum > College/University Physics Help > Advanced Thermodynamics

Advanced Thermodynamics Advanced Thermodynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Jan 14th 2016, 01:20 PM   #1
Junior Member
 
Join Date: Jan 2016
Posts: 3
defining entropy with heat instead of work?

From what I understand, in the Carnot cycle summing qi/Ti for each step results in zero, thus indicating a new state function, entropy = qrev/T. But since dE = 0 = q+w, then q = -w, and looking at the equations derived from the cycle summing wi/Ti for each step should also result in zero. So why can't one also define entropy as wrev/T?
tabby123 is offline   Reply With Quote
Old Jan 14th 2016, 01:37 PM   #2
Senior Member
 
Join Date: Apr 2015
Location: Somerset, England
Posts: 987
I don't like using the Carnot cycle to introduce entropy as this property appears in other circumstances, even when no work is done.

However, what does a Carnot machine do?

Is it 100% efficient at this?

Answer this and you will have the Clausius statement of the second law and the answer to your own question.

It is impossible in a continuous cycle to transfer heat from a colder to a warmer body without net changes in other bodies.
studiot is offline   Reply With Quote
Old Feb 23rd 2016, 11:19 PM   #3
Junior Member
 
Join Date: Jan 2016
Posts: 3
reversible vs irreversible work for adiabatic process

Hello, I have another question:

I have a gas transitioning adiabatically between A (P1, V1) and B (P2, V2) where P1>P2 and V2>V1. The question is to determine the net work done on the gas if the gas is first expanded reversibly from A to B (w = dE = Cv(T2-T1)), and then compressed irreversibly from B to A (w = -Pext(V1-V2)) at a constant external pressure defined by A. In this scenario, simply looking at the areas under the graphs the net work should be positive.

I am trying to reconcile this with dE for the gas. For the roundtrip transition (A to B to A), dE = 0. And if we take each step as adiabatic, then dE = w for each step, but as I have described above you would end up with two different values for dE for each step, thus dE not equal to zero. My logic is flawed somewhere. If I compress irreversibly would the transition still be adiabatic? Alternatively, is the original scenario flawed: can I have an irreversible adiabatic transition?
tabby123 is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Advanced Thermodynamics

Tags
defining, entropy, heat, thermodynamics, work



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Thermodynamis Help Please!! Heat flow, work done, etc. mmmboh Thermodynamics and Fluid Mechanics 7 Apr 11th 2017 10:16 AM
How can increased entropy accompany decreased heat capacity? adamjohnson Thermodynamics and Fluid Mechanics 1 Sep 6th 2016 08:44 PM
Heat and Work Thermodinamic veterman Thermodynamics and Fluid Mechanics 1 Apr 18th 2011 12:04 PM
please help in defining disturbance eason123deng Waves and Sound 0 Jan 5th 2011 09:12 AM
work and heat ...please help me, i really don't get these problems.. mahamehdi Thermodynamics and Fluid Mechanics 2 Feb 21st 2009 11:59 AM


Facebook Twitter Google+ RSS Feed