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

Answer:

1.

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

2. $W=-8.2kPa$

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:

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

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

$P_1V_1^n=P_2V_2^n\\\\\frac{P_1}{P_2}=(\frac{V_2}{V_1} )^n\\\\\frac{1bar}{2bar}= (\frac{0.04m^3}{0.1m^3} )^n\\\\0.5=0.4^n\\\\n=\frac{ln(0.5)}{ln(0.4)} =0.7565$

Now, we compute the work:

$W=\frac{P_2V_2-P_1V_1}{1-n} =\frac{2bar*0.04m^3-1bar*0.1m^3}{1-0.7565} \\\\W=-0.082bar*m^3*\frac{1x10^2kPa}{1bar}\\ \\W=-8.2kPa$

Regards.