- #1

George32

- 7

- 0

## Homework Statement

We have a piston in a cylinder containing Helium. The initial states are:

P

_{1}=150kPa

T

_{1}= 20°C

V

_{1}=0.5m

^{3}

Following a polytropic process, the final states are:

P

_{2}=400kPa

T

_{2}=140°C

V

_{2}is unknown.

We're also given R=2.0769 kPa.m

^{3}.(kg.K)

^{-1}

And C

_{p}=5.1926kJ.(kg.K)

^{-1}

As well as the temperature of the surrounding being 20°C

We are then asked to find:

a) The entropy change inside the system

b) The entropy change in the surroundings

c) Is the process Irreversible, Reversible or Impossible?

A hint is that we need to find n (the polytropic constant)

## Homework Equations

PV=RT (1)

[itex]TdS=dh - VdP[/itex] (2)

[itex]dh=CpdT[/itex] (3)

[itex]PV^n = Constant[/itex] (4)

## The Attempt at a Solution

I have a solution for part (a) which comes out at -0.2546kJ/K.

I got this after rearranging (1) to give [tex]\frac{V}{T} = \frac{R}{P} [/tex] and filling (3) and the new (1) into (2) to give [tex] ds=\frac{CpdT}{T} - \frac{RdP}{P}[/tex]

Integrating each term between their respective limits to give [tex]\Delta s = Cp(ln (\frac{T2}{T1})) - R(ln(\frac{P2}{P1}))[/tex]

Now for the process to be polytropic it has to be quasi-static which doesn't necessarily mean reversible but is (as I understand it) usually reversible, so the entropy change of the surroundings would have to be 0.2546 at least. However I don't know how I would calculate this, and the third part makes me question if I should get an answer lower and it will be an impossible process.

Any pointers on how to proceed, as well as perhaps a confirmation of using the correct process so far, would be great, thanks :)

G