# Reversible adiabatic expansion of ideal gas, entropy change?

1. Jan 31, 2010

### 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"

Khris

2. Jan 31, 2010

### Andrew Mason

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

3. Jan 31, 2010

### krhisjun

i know
$$\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T$$
but dQ is zero for an adiabatic process isnt it?

Khris

4. Nov 15, 2010

### bon

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

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

5. Nov 16, 2010

### bon

heloooo? *bump*