December 10, 2024, 01:05:56 AM
Forum Rules: Read This Before Posting


Topic: Question on second law of thermodynamic  (Read 11205 times)

0 Members and 1 Guest are viewing this topic.

san1984

  • Guest
Question on second law of thermodynamic
« on: April 16, 2006, 10:37:31 AM »
The question goes like this. A fundamental fact of experience on which the second law of thermodynamics is based that the heat absorbed at any one temperature cannot be completely transformed in work without leaving some change in the system or its surroundings.


Consider now the following. When an ideal gas is allowed to expand isothemally against a piston, delta U = q + w =0. Thus in this case the work done by the gas is equal to the heat transferred from the reservoir to the gas, and the efficiency of turning into work is 100%. Explain why this is not a violation of the Second Law ?   ???

Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 785
  • Mole Snacks: +77/-37
  • Gender: Male
Re: Question on second law of thermodynamic
« Reply #1 on: April 17, 2006, 08:27:56 AM »
http://en.wikipedia.org/wiki/Isothermal

according to it, since dU = dq + dw,

and dq = 0 (constant temperature)

dU = dw

and hence the change in internal energy must be converted entirely into work 
one learns best by teaching

san1984

  • Guest
Re: Question on second law of thermodynamic
« Reply #2 on: April 17, 2006, 10:23:15 AM »
hmm... I think that being an isothermal process, q is not 0 in this case but instead change in internal energy is equal to zero and hence q = w.

But the main part of the question is still asking for why it is not a violation of the second law of thermodynamics in this scenerio.

Offline Winga

  • Chemist
  • Full Member
  • *
  • Posts: 510
  • Mole Snacks: +39/-10
Re: Question on second law of thermodynamic
« Reply #3 on: April 17, 2006, 06:53:00 PM »
The system becomes much random.

Offline Hunt

  • Chemist
  • Full Member
  • *
  • Posts: 240
  • Mole Snacks: +25/-7
  • Gender: Male
Re: Question on second law of thermodynamic
« Reply #4 on: April 17, 2006, 08:17:54 PM »
I think you're right,  san1984, U = 3/2 RT for a gas, at least from KMT perspective. Perhaps this is the case because we're dealing with an ideal gas here?

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Question on second law of thermodynamic
« Reply #5 on: April 17, 2006, 08:51:19 PM »
The interpretation of the second law given in the OP applies only to engine cycles (i.e. thermodynamic processes which start and end at the same thermodynamic states).  Since an isothermal expansion is not a cyclic process, you can't apply this interpretation of the second law.

san1984

  • Guest
Re: Question on second law of thermodynamic
« Reply #6 on: April 18, 2006, 01:23:28 AM »
oic... now it makes more sense now..  :D
tks !

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3177
  • Mole Snacks: +261/-13
  • Gender: Male
    • New Asia Republic
Re: Question on second law of thermodynamic
« Reply #7 on: April 18, 2006, 04:15:19 AM »
du = q + w
du = Cv.dT
q = T.ds
w = - p.dv
Cv.dT = T.ds - p.dV

Since this is an isothermal process, then dT = 0
T.ds = p.dv
ds = (p/T)dv

Since we are dealing with a perfect gas, then p/T = R/V
ds = (p/T)dv = (R/V)dv
=> ?S = R.ln(Vfinal/Vinitial)

Since we are dealing with an expansion process, then Vfinal > Vinitial
=> Vfinal/Vinitial > 1
=> ln(Vfinal/Vinitial) > 0
=> ?S > 0

The second law is not violated at all.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Hunt

  • Chemist
  • Full Member
  • *
  • Posts: 240
  • Mole Snacks: +25/-7
  • Gender: Male
Re: Question on second law of thermodynamic
« Reply #8 on: April 18, 2006, 07:51:57 AM »
Quote
The interpretation of the second law given in the OP applies only to engine cycles (i.e. thermodynamic processes which start and end at the same thermodynamic states).  Since an isothermal expansion is not a cyclic process, you can't apply this interpretation of the second law.

It makes sense now, but just out of curiousity , what exactly is the internal energy of a real gas? Is it a function of temperaure only?

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3177
  • Mole Snacks: +261/-13
  • Gender: Male
    • New Asia Republic
Re: Question on second law of thermodynamic
« Reply #9 on: April 18, 2006, 09:48:09 AM »
The internal energy of an ideal gas is a function of its temperature.

However, this is not true for real gases. This is reflected in the various equations of state.

The internal energy for the Van Der Waals' Equation of State:
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Hunt

  • Chemist
  • Full Member
  • *
  • Posts: 240
  • Mole Snacks: +25/-7
  • Gender: Male
Re: Question on second law of thermodynamic
« Reply #10 on: April 18, 2006, 12:23:43 PM »
Thanks , Geodome

Sponsored Links