reversible adiabatic expansion of ideal gas, entropy change?

krhisjun

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

Andrew Mason

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

krhisjun

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

bon

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

Surely here dQ = 0 since the gas is expanded adiabatically?

bon

heloooo? *bump*