Question

In an industrial process, a fuel-air mixture is fitted with a piston. When the mixture is ignited at a constant pressure of 700mm Hg, 2.0 kJ of energy is released and is found to occupy a final volume of 200L. What would be the initial volume occupied by the mixture if all energy released is converted as work done by the system to push the piston?

Answer 1

Given: Pressure of the system = 700 mm Hg   = 93.3257 KPa

Energy released = 2.0 KJ

Final volume occupied by the mixture when it is ignited (V1) = 200 L = 0.2 m3

To find:   initial volume occupied by the mixture (before ignition) = V2

solution:

Formula used:  U = - W

where, U is the change of internal energy of the system

W is the work done by the system = P * V

2 = - 93.3257 * ( V2- V1)

   2 = - 93.3257 * ( V2 - 0.2)

V2 = 0.178569 m3

or V2 = 178.569 L

So, initial volume occupied by the mixture (before ignition) = 178.569 L