- #1

hinjab

## Homework Statement

- A container of 1.5 Kg of gas is at a temperature and pressure of 293 K and 1 bar respectively. The gas is adiabatically compressed until its temperature and pressure are 450 K, 4.49 bars. Adiabatic processes are processes with no heat transfer. The properties of this gas are cv = 10.3 KJ/(Kg K) and R = 4.158 KJ/(Kg K). Neglect kinetic and potential energy terms.

## Homework Equations

a) Use the first law to determine the work into the system.

b) Calculate the entropy production for this process.

c) Is this a reversible process?

## The Attempt at a Solution

I solved A by saying that the work done = the change in internal energy. So Work done = P1V1 - P2V2 / Y-1 where Y = Cp/Cv = 1.403.

Since P1V1 = NRT1 and P2V2 = nRT2 , I calculated Work done as nRT1-nRT2 / 1.403 - 1 = -2427.79 kJ. The negative sign representing work being done on the system.

I am having trouble calculating B. My understanding would be that the change in entropy for an adiabatic process where volume is constant would be calculated as Cv ln (T2/T1), However I am finding conflicting ideas online, where people are actually calculating it as Cp ln (T2/T1) - R ln (P2/P1). When I calculate it that way, I get a negative number of -61.71 J/K for entropy production.

Can anyone explain if that is the right way to calculate it, or if I should be doing it the way I originalyl thought was correct (Entropy production = Cv ln (T2/T1)?Thanks!