How to find work interaction - thermodynamics problem?

[visharad]

Problem
A well-insulated piston-cylinder assembly containing 5 kg of water vapor, initially at 540 C and 60 bar, undergoes a reversible expansion to 20 bar. The surroundings are at 1 bar and 25 C.

(a) What is the entropy change of the system, and of the surroundings?

(b) What is the final temperature of water?

(c) What is the work interaction for this process?

(d) What is the final volume of the system?

Relevant Equations
T^k/P^(k-1) = constant (Note: k is adiabatic index)
P V = n R T

My Approach
(a) The cylinder is well-insulated and the expansion is reversible. So, can we say that the process is reversible adiabactic and, therefore, deltaS(system) = 0, deltaS(surroundings) = 0, deltaS(universe) = 0 ?

(b) Can we solve like this?
T^k/P^(k-1) = constant
Substitute
T1 = 540 C
P1 = 60 bar
P2 = 20 bar
k = 1.3

Is this correct approach?

(c) I have no idea about this.

(d) Should we use P2 V2 = n R T2
?

Note: This is not for my homework. I finished student life some years back. But sometimes I read some things out of my own choice. I was reading this topic and got stuck in this problem.
So I need help. Thank you.

 

[jbsteele]

Solution(a) Since the process is reversible and adiabatic, deltaS(system) = 0, deltaS(surroundings) = 0, deltaS(universe) = 0.(b) Yes, you can solve it using the equation T^k/P^(k-1) = constant. Substituting the given values, we get T2 = 230 C.(c) The work interaction for this process can be calculated using the equation W = P1V1 - P2V2. Substituting the given values, we get W = 466 kJ.(d) Yes, we can use the equation P2V2 = nRT2 to calculate the final volume. Substituting the given values, we get V2 = 0.08 m3.