Question

Question

In an experiment a gas was found to obey the following equation of state:


(P + 10^3)V = 2.5 x 10^4 J

Find the work done when the gas expands from 1 m3 to 3 m3.

_______

the answer is suppose to be

2.55 x 104J

____________ what i did so far


2 ways i approached this question...first way was the use V as 2 (that is Vf- Vi) and solve for P

Then i plug in equation

W = p delta V

got 2.7 X 10^4


____________ nxt way

i found the values of p using V as 1 and 3

Then, i found the work for each of the pressure i calculated, then added...got like 3.8 x 10^4

i'm baffled

Answer 1

The work done is the integral of P dV, with P a function of V.

That is the integral of
[2.5 *10^4/V - 10^3 ]dV
from V=1 to V=3

= 2.5*10^4 ln 3 - 1000 (3-1)
= 25,500 J

Answer 2

2.5 *10^4/V - 10^3 ]dV
from V=1 to V=3


DOn't GET HERE

Answer 3

[2.5 *10^4/V - 10^3 ]dV
is what the P dV integral is. The expression in brackets is P.
Do the algebra.

Integral PdV is the definition of work.