Question

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given by p = AV2 + BV-2 (p is in bar and V is in m3). The initial volume is 0.1 m3, and the initial pressure is 1 bar. The final volume is 0.04 m3, and the final pressure is 2 bar. Determine:

Answer 1

1.

`A=69.8(bar)/(m^6)\\\\ B=0.00302bar*m^6`

2.
`W=-8.2kPa`

Step-by-step explanation:

Hello,

1. In this case, for the given p-V equation, one could use the two states to form a 2x2 linear system of equations in terms of A and B:

`\left \{ {{0.1^2A+0.1^(-2)B=1} \atop {0.04^2A+0.04^(-2)B=2}} \right.`

`\left \{ {{0.01A+100B=1} \atop {0.0016A+625B=2}} \right`

Whose solution by any method for solving 2x2 linear system of equations (elimination, reduction or substitution) is:

`A=69.8(bar)/(m^6)\\\\ B=0.00302bar*m^6`

2. Now, for us to compute the work, we must first compute n, as the power relating the pressure and volume for this process:

`P_1V_1^n=P_2V_2^n\\\\(P_1)/(P_2)=((V_2)/(V_1) )^n\\\\(1bar)/(2bar)= ((0.04m^3)/(0.1m^3) )^n\\\\0.5=0.4^n\\\\n=(ln(0.5))/(ln(0.4)) =0.7565`

Now, we compute the work:

`W=(P_2V_2-P_1V_1)/(1-n) =(2bar*0.04m^3-1bar*0.1m^3)/(1-0.7565) \\\\W=-0.082bar*m^3*(1x10^2kPa)/(1bar)\\ \\W=-8.2kPa`

Regards.