Question

A mass of gas at an initial internal energy of 300 kJ is allowed to expand behind a piston untill it's internal energy is 200 kJ, the value of final pressure and volume are 4.59 bar and 0.148 m^3 respectively, the law of expansion is PV^2 =C calculate

1) work done

2) initial volume.

Answer 1

Solution:

1) From the first law of thermodynamic: 

Q=ΔU+W

W=Q-ΔU

W=100kJ

2)

100 kJ = integral from V₁ to V₂ of p dV:

$(4.59 bar)(0.148m^3)[\frac{1}{V_1}-\frac{1}{V_2}]=>\frac{100kJ}{(459000Pa)(0.148m^3)}=\frac{0.148m^3}{V_1-1}$

$2.472=\frac{0.148m^3}{V_1}$

$V_1=0.006m^3$