Thermodynamics: Isobutane entropy and power problem

[thermohelp]

Homework Statement

Isobutane flows through an expander (like a turbine) at 680 moles/sec where it undergoes a change in state from 4826 kPa and 260 degrees Celsius to 483 kPa and 195 degrees Celsius.

a) If the expander is adiabatic how much power is generated by the expander?
b) What is the change in entropy for this operation?

Homework Equations

T_C = 408.1 K
P_C = 3650 kPa
C^ig_R = A + BT + C(T^2) + D/(T^2) where T is in K, A = 1.667, B = 37.853*10^-3, C = 11.945*10^-6, and D = 0.

The Attempt at a Solution

I have no idea where to go with this problem for parts A and B. Can someone please give me a way to go about solving this problem

 

[BigJDubs]

?For part A, you need to use the first law of thermodynamics. It states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). Since this is an adiabatic process, no heat is added (Q=0) so ΔU=W. The power generated by the expander will be equal to the rate of change of internal energy with respect to time, or P=ΔU/Δt. You can calculate the change in internal energy by using the ideal gas equation of state. The change in internal energy can be calculated as follows:ΔU = nRT(1/P1 - 1/P2)where n is the number of moles of gas, R is the universal gas constant, T is the temperature and P1 and P2 are the initial and final pressures, respectively. Therefore, the power generated by the expander is:P = nRT(1/P1 - 1/P2)/ΔtFor part B, you need to use the second law of thermodynamics. It states that the change in entropy (ΔS) is equal to the heat added to the system (Q) divided by the temperature (T). Since this is an adiabatic process, no heat is added (Q=0) so ΔS=0. Therefore, the change in entropy for this operation is zero.