Recent content by erick

W = ∫ P dV = P ∫ dV = p(V2 – V1) what kind of effort do you want me to do? i'll try to solve it but still I'm wrong..

and tnx for nothing

ok..

i know that formula...

thermodynamics sir.

hmmm, is that the correct formula?

/sad. ok, I'm out. i can't solve it, but thnx for the help. goodluck..

Wnf = pV lnP1/P2 = (35 N / m²)(0.12 m³)[ln (35 kPa / 80 kPa = -3.4692 x 10ˉ³ N-m = -3.4692 kJ Q = ΔU + W = 3.22 kJ + (-3.47 kJ) = - 0.25 kJ

weee, I'm done with #1 but i don't know my answer if it's right. = -0.25 kJ

I'm having trouble solving... I've try a lot but still can't get it. Could anyone write the solutions and with steps on how they did it so I can possibly learn! Thanks a lot

1. Ethane, initially at 35 kPa with a volume of .12 m³, is compressed without friction in a cylinder until its volume is halved in such a manner that its pressure and volume are linearly related, p = a + bV. The final pressure is 80 kPa and the change in internal energy of the ethane is 3.22...