Question

In an isobaric process, volume of a system was changed from 10 m3 to 2 m3 at 200 kPa. The work done, W, in this process based on the sign convention in the class is

-1600 kJ

1600 kJ

-25 kJ

25 kJ

Answer 1

The concept required to solve this problem is that related to the Isobaric process. Isobaric process is understood as the process in which changes occur at constant pressure. From the first law of thermodynamics this can be expressed as,

`W = \int P dV`

Here,

P = Pressure

dV = Differential of Volume

As the Pressure is constant we have,

`W = P \int dV`

`W = P \big [ V \big ]^(V_f)_(V_i)`

`W = P (V_f-V_i)`

Replacing

`W = 200*10^3 (2-10)`

`W = -1600kJ`

Therefore the correct answer is A.