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Entropy change question

  • Thread starter henryc09
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  • #1
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Homework Statement


(a) A piston is used to compress an ideal gas quasistatically from volume Vi to volume Vf . If the gas is in thermal contact with a heat bath at temperature T, such that the compression is carried out isothermally, calculate the work done on the gas, the change in entropy of the gas and the change in entropy of the heat bath
(b) The compression is repeated but but non-quasistatically. Are the three calculated quantities higher, lower, or the same as before?


Homework Equations




The Attempt at a Solution



For (a) I said that W=-NkT*ln(Vf/Vi), and [tex]\Delta[/tex]E = 0, so [tex]\Delta[/tex]Q = NkT*ln(Vf/Vi).

Then as it's quasistatic [tex]\Delta[/tex]Sgas = [tex]\Delta[/tex]Q/T = Nk*ln(Vf/Vi)

I then said as it's quasistatic [tex]\Delta[/tex]Suniverse=[tex]\Delta[/tex]Sgas + [tex]\Delta[/tex]Sbath=0
so [tex]\Delta[/tex]Sbath = -Nk*ln(Vf/Vi)

I think this is right but confirmation would be good. For (b) I said that [tex]\Delta[/tex]Sgas will be the same as it's a state variable, but as the entropy of the universe must increase the bath's entropy must increase. The only thing I'm unsure about is whether or not the work increases/decreases/stays the same. Anyone got any ideas about how to approach this?
 

Answers and Replies

  • #2
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hi
why dont you look at the part about joule expansion in the blundell book?
it has an example that is similar-ish, but im nt sure whether it givs the right answer for a
as for b and c and d, i hav no idea
now do u know how to do the second part of que 2b on the prob sheet? if so, please reply to the thread i started
 
  • #3
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I don't have the Blundell book, and I haven't got that far yet no.

I think I've figured this out though, but again confirmation would be good:

work will be the same as [tex]\Delta[/tex]Sgas is the same therefore [tex]\Delta[/tex]Q is the same. As T is constant [tex]\Delta[/tex]E is 0 so [tex]\Delta[/tex]Q=-[tex]\Delta[/tex]W as before.
 
  • #4
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i dunno
im too tired to think
but i looked at what someone else did and they agree with the textbook when they hav the entropy change of gas= -entropy change of heat bath, but i dunno what we are supposed to express the entropy in. the textbook expresses it in terms of R and stuff but someone else got it in terms of kT or something

do you know what difference it makes whether the compression is quasistatic or not?
 
  • #5
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i've heard ppl say that the change in entropy for que1b would be higher because of the dSi (dS subcript i) term in the equation for a non-quasistatic process:

dS=(dQ/Tr)+dSi

but im not sure
does anyone know for sure please?
 
  • #6
Andrew Mason
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Homework Helper
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The only thing I'm unsure about is whether or not the work increases/decreases/stays the same. Anyone got any ideas about how to approach this?
If the compression occurs very quickly what happens to the temperature of the gas? What effect does that have on pressure and the work that must be done on the gas to compress the gas?

AM
 
  • #7
241
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but if the pressure increases but volume decreases and W=pdV then what difference does it make to the work done?
 

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