Reversible adiabatic expansion of ideal gas, entropy change?

[krhisjun]

Homework Statement

One mole of an ideal gas at 0 celsius 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 doesn't state this i don't feel that i can just state "assuming system is isolated entropy change is zero"
so I am stuck how to go about this..

Khris

 

[Andrew Mason]

Homework Statement

One mole of an ideal gas at 0 celsius 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 doesn't state this i don't feel that i can just state "assuming system is isolated entropy change is zero"
so I am 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).

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

AM

 

[krhisjun]

i know
<br /> \Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T<br />
but dQ is zero for an adiabatic process isn't it?

Khris

 

[bon]

The entropy change of the gas is not zero.

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

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

AM

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

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

 

[bon]

The entropy change of the gas is not zero.

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

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

AM

heloooo? *bump*