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A question on second law of thermodynamics

  • Thread starter ky92
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  • #1
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Homework Statement


calculate the change of entropy (for the system) when 3 moles of a monatomic perfect gas, for which Cp,m = (5/2) R, is heated and compressed from 298K and 1atm to 398K and 5atm.

ans: -22.1 JK-1

Homework Equations


for an ideal gas, Pv=nRT
for iso-choric condition, delta S = Cv ln(T2/T1)

The Attempt at a Solution


as both the P and T increase, by Pv=nRT v should be constant at this case
so i thought i should use Cv ln(T2/T1)

however the model answer is that it uses nCp,m ln(T2/T1) + nRln(V2/V1)

i dont know which part of my concept is wrong...
pls help, thank you!
 

Answers and Replies

  • #2
Mapes
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Hi ky92, welcome to PF. Just because P and T are changing doesn't mean that V will remain constant! All three could change simultaneously. When this happens, it's common to simplify the problem by assuming that two simpler processes (isobaric and isothermal, for example) occur in sequence to produce the desired final state.
 
  • #3
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thank you!

by the way
why couldn't i replace (V2/V1) by (T2/T1) in this case? (but we can replace V2/V1 by P1/P2)
i dont really get it!

how to decide whether we should substitute between P-V or P-T/V-T?
 
  • #4
Mapes
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In an isothermal process on an ideal gas, [itex]V_2/V_1=P_1/P_2\neq T_1/T_2\left(=1\right)[/itex]. You can show this with the ideal gas law.
 
  • #5
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In an isothermal process on an ideal gas, [itex]V_2/V_1=P_1/P_2\neq T_1/T_2\left(=1\right)[/itex]. You can show this with the ideal gas law.
oh i get it now
it's just because we assume the reaction is an isobaric, then isothermal process
so T is constant for the second part i.e. in nR ln(V2/V1)

thanks!
you help me a lot
i am bad in physics :P
 
  • #6
Mapes
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You got it. Hey, nobody's born knowing thermodynamics; keep at it.
 

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