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Reversible adiabatic expansion of ideal gas, entropy change? 
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#1
Jan3110, 01:30 PM

P: 22

1. The problem statement, all variables and given/known data
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] 2. Relevant equations I know delta Q = nRT Ln (Vf/Vi) for a reversible isothermal expansion delta S = delta Q / absolute T dE = dW + dQ 3. 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 


#2
Jan3110, 02:18 PM

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P: 6,654

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
Jan3110, 02:32 PM

P: 22

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
Nov1510, 01:12 PM

P: 567

Reversible adiabatic expansion of ideal gas, entropy change?
Surely here dQ = 0 since the gas is expanded adiabatically? 


#5
Nov1610, 09:48 AM

P: 567




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