Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state (a cycle).

Process 1-2: Expansion from State 1 where p1 = 10 bar, V1 = 1 m3 , to State 2 where V2 = 4 m3 . During the process, pressure and volume are related by pV1.5 = constant.
Process 2-3: Constant volume heating to State 3 where p3 = 10 bar.
Process 3-1: Constant pressure compression to State

Sketch the processes on p-V coordinates and evaluate the work for each process, in kJ. What is net work for the cycle, in kJ?

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given  

Process 1 ---> 2

The relation of the process P = constant

Pressure of point (1) P1 =  10 bar = P2

Volume of point (1) V1   = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant  

Process 2 ---> 3

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

Process 3 ---> 1

The relation of the process PV = constant  

Required  

Sketch the processes on the PV coordinates

The work for each process in kJ  

Solution  

The work is defined by  

W=

a=V2

b=V1

x=P

dx=dV

Process 1 ---> 2  

P3 = P4 = 5 bar  

W=

a=V3

b=V2

x=4

dx=dV

putting the value of a, b, x, dx in above integral

W=400 kJ

Process 2 ---> 3

V = constant Then there is no change in the volume,hence W = 0 kJ  

Process 3 ---> 1  

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1  

 W=

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

  = -160.944 KJ

i think its

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.