Question

A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 = 10 bar, V1 = 0.1 m3 and p2 = 1 bar, V2 = 1.0 m3: Process A: Process from 1 to 2 during which the pressure-volume relation is pV = constant. Process B: Constant-volume process from state 1 to a pressure of 2 bar, followed by a linear pressure-volume process to state 2. Kinetic and potential energy effects can be ignored.
For each of the processes A and B, (a) sketch the process on p-V coordinates, (b) evaluate the work done by the gas, in kJ.

Answer 1

Answer:

a) 90 kJ

b) 230.26 kJ

Explanation:

The pressure at the first point  $P_{1}$= 10 bar —> 10 x 102 = 1020 kPa

The volume at the first point  $V_{1}$= 0.1 m^3  

The pressure at the second point $P_{2}$= 1 bar —> 1 x 102 = 102 kPa

The volume at the second point $V_{2}$ = 1 m^3  

Process A.

constant volume V = C from point (1) to P = 10 bar.

Constant pressure P = C to the point (2).  

Process B.

The relation of the process is PV = C  

Required  

For process A & B

(a) Sketch the process on P-V coordinates

(b) Evaluate the work W in kJ.  

Assumption  

Quasi-equilibrium process

Kinetic and potential effect can be ignored.  

Solution

For process A.

V=C  

There is no change in volume then

$W_{a(1)}= 0\\P=10^{2}$

The work is defined by  

$W_{a(2)}=\int\limits^V_V {P} \, dV$

$W_{A(2)} =$║$10^{2}$ V║limit 1--0.1

$W_{A(2)} =$ 90 kJ

Process B  

PV=C  

By substituting with point (1) C = 10^2 x 1= 10^2  

The work is defined by

$W_{b}=\int\limits^V_V {P} \, dV\\P=10^{2} V^{-1}\\ W_{b}=\int\limits^V_V {10^{2} V^{-1}} \, dV$

$W_{A(2)} =$ ║$10^{2}$ ln(V)║limit 1--0.1

         =230.26 kJ