The Entropy Change Argon in a non-isothermal process expansion.

  • #1

Homework Statement


a) Calculate the entropy change when Argon, at 25oC and 0.101 MPa pressure, in
a container of volume 0.5 litre is allowed to expand to 1.0 litre and is simultaneously
heated to 100oC.
b) Calculate the entropy change when the same initial sample is compressed to 0.05
litre and cooled to –25oC.
(Hint: choose a suitable path or combination of successive paths to move between the
two states described )



Homework Equations


Delta S =nC[tex]_{}V[/tex]ln(T[tex]_{}2[/tex]-T[tex]_{}1[/tex]) + nRln(V[tex]_{}2[/tex]/V[tex]_{}1[/tex])

Where since Argon is monotonic C[tex]_{}V[/tex]=3R/2



The Attempt at a Solution


But the question does not state the value of n and since this is not an ideal gas I cannot work it out.
 

Answers and Replies

  • #2
97
1
Oh, c'mon :P. If you are allowed to assume, that argon is an ideal gas, and pV=nRT applies, read your first sentence again. Y~ou've got temperature, pressure and volume. You might want to use it to calculate n...
 
  • #3
Except you get different values of n at the two different values of temperature and corresponding volume.
So you cant assume ideality...
 
  • #4
Mapes
Science Advisor
Homework Helper
Gold Member
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There is only one state where you are given pressure, temperature, and volume: the initial state.
 
  • #5
97
1
But it doesn't say that the pressure remains constant, or at least you didn't write it.
You've got: p1=0.101 MPa, T1=25C, V1=0.5l and T2=100C, V2=1l. You need to assume that the pressure has changed.

Edit: Dammit, Mapes was faster :P.
 
  • #6
Ohhhh It's suddenly clicked!
I was assuming it was isobaric even though the question doesnt actually suggest it is...
Wow I feel stupid =p
Cheers for your help
 

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