Thermo dynamics, entropy based problem

MightyG

1. The problem statement, all variables and given/known data

A rigid vessel of total volume 0.5m3 contains 1kg of air at 15°C initially
constrained in a volume of 0.06m3 by a diaphragm. The remaining volume is
completely evacuated (pressure = 0). If the diaphragm is burst, allowing free
adiabatic expansion of the air, determine the change of entropy of the air and
of the surroundings. If instead of a free expansion, the process had been a
reversible expansion between the two states, what would then be the change
of entropy of the air, and of the surroundings? Compare the net change of
entropy for the system and surroundings in the two cases. Surrounding
temperature 15°C .
Answer: +0.609, 0, +0.609 – 0.609 kJ/K

2. Relevant equations

[tex]\Delta[/tex]S=C_v.Ln(P2/P1) + C_p.Ln(V2/V1) (1)

[tex]\Delta[/tex]S=C_v.Ln(T2/T1) + R.Ln(V2/V1) (2)

[tex]\Delta[/tex]S=C_p.Ln(T2/T1) - R.Ln(P2/P1) (3)

[tex]\Delta[/tex]S=C_v.Ln(P2/P1) - C_p.Ln(V2/V1) (4)

3. The attempt at a solution

I can work out the first part of the problem easy enough taking the expansion as being isothermic so (2) becomes [tex]\Delta[/tex]S=R.Ln(V2/V1) but im stumpped by the second part.

im told that the Eq becomes;

Q = [tex]\Delta[/tex]u + W since there is no heat transfer to the system it becomes Q=W=m.R.t.Ln(V2/V1) but ive got no idea where this Eq comes from.#

EDIT:

Is this just a case of W=[tex]\int[/tex]PdV and since Pv=mRt so P=(mRt)/v

then W=[tex]\int[/tex][tex]\frac{mRt}{V}[/tex] dV

= mRt[tex]\int[/tex][tex]\frac{1}{V}[/tex] dV

MightyG

ok, just worked that through and I get the right magnitude for the entropy change but I dont understand why the answer is negative.

my working:

[tex]W = mRt\int^{2}_{1}\frac{1}{v}dv = mRt [Ln v]^{2}_{1} = 1*0.287*288*(Ln0.5 - Ln0.06) =175.253[/tex]

so,

[tex]\Delta S = \frac{Q}{T} = \frac{175.253}{288} = 0.609Kj[/tex]

but acording to the answers it should be -0.609Kj?