Reversible adiabatic expansion of ideal gas, entropy change?

AI Thread Summary
The discussion centers on calculating the change in entropy for an ideal gas undergoing reversible adiabatic expansion. Participants confirm that for an isothermal expansion, the entropy change is 5.76 J/K, while the entropy change for the adiabatic process is debated. It is noted that dQ equals zero in an adiabatic process, leading to confusion about whether the entropy change can be considered zero. However, the consensus is that the entropy change is not zero, and hints are provided to approach the calculation without integration. The conversation emphasizes the importance of understanding entropy changes in different thermodynamic processes.
krhisjun
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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 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
 
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krhisjun said:

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 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
 
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
 
Andrew Mason said:
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?
 
Andrew Mason said:
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*
 
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