Calculate the work done by the torque.

Explanation

(a) The pressure `"P = 1 atm = 101 kPa"`
`V_(f)=2m^(3) and V_(i)=1m^(3)`
From equation `W=int_(V_(i))^(V_(f))PdV=Pint_(V_(i))^(V_(f))dV`
Since P is constant. It is taken out of the integral.
`W=P(V_(f)-V_(i))=101xx10^(3)xx(2-1)=101kJ`
(b) Since the pressure is kept constant, PV diagram is straight line. The area is equal to work done by the gas.
Note the arrow mark in the curve. Suppose the work is done on the system, then volume will decreases and the arrow will point in the opposite direction.