Reversible adiabatic expansion of ideal gas, entropy change?

  • Thread starter krhisjun
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  • #1
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Homework Statement


One mole of an ideal gas at 0 celcius is subjected to changes below, calculate the change in entropy of the gas:

i) Gas is expanded reversibly and isothermally to twice its initial volume. DONE - 5.76 J/K
ii)A similar expansion to i. is performed reversibly and adiabatically [hint, for an ideeal gas undergoing reversible adiabatic expansion TV^(gamma - 1) = constant]


Homework Equations



I know delta Q = -nRT Ln (Vf/Vi) for a reversible isothermal expansion

delta S = delta Q / absolute T

dE = dW + dQ

The Attempt at a Solution



for a isolated system i know the entropy change would be zero, but as it doesnt state this i dont feel that i can just state "assuming system is isolated entropy change is zero"
so im stuck how to go about this..

Khris
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


One mole of an ideal gas at 0 celcius is subjected to changes below, calculate the change in entropy of the gas:

i) Gas is expanded reversibly and isothermally to twice its initial volume. DONE - 5.76 J/K
ii)A similar expansion to i. is performed reversibly and adiabatically [hint, for an ideeal gas undergoing reversible adiabatic expansion TV^(gamma - 1) = constant]


Homework Equations



I know delta Q = -nRT Ln (Vf/Vi) for a reversible isothermal expansion

delta S = delta Q / absolute T

dE = dW + dQ

The Attempt at a Solution



for a isolated system i know the entropy change would be zero, but as it doesnt state this i dont feel that i can just state "assuming system is isolated entropy change is zero"
so im stuck how to go about this..
The entropy change of the gas is not zero.

Work it out (Hint: you don't actually have to do any integration).

[tex]\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T[/tex]

AM
 
  • #3
22
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i know
[tex]
\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T
[/tex]
but dQ is zero for an adiabatic process isnt it?

Khris
 
  • #4
bon
559
0
The entropy change of the gas is not zero.

Work it out (Hint: you don't actually have to do any integration).

[tex]\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T[/tex]

AM

Sorry to bring this up again, but doing a similar question...

Surely here dQ = 0 since the gas is expanded adiabatically?
 
  • #5
bon
559
0
The entropy change of the gas is not zero.

Work it out (Hint: you don't actually have to do any integration).

[tex]\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T[/tex]

AM

heloooo? *bump*
 

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