Work done ON the fluid is determined by taking negative ofthe area under given curve from Point (i) to Point (f):

Work Done ONFluid For (i) ---> (f)} = -{Area Under Curve} =

\(\displaystyle=-{\left({6.0}{e}+{6}\right)}\cdot{\left({1}\right)}\pm{\left({\frac{{{1}}}{{{2}}}}\right)}\cdot{\left({6.0}{e}+{6}+{2.0}{e}+{6}\right)}\cdot{\left({1}\right)}\pm{\left({2.0}{e}+{6}\right)}\cdot{\left({1}\right)}\)

\(\displaystyle=-{1.20}{e}+{7}\) Joules

Work done ON fluid during reverse compression along same pathfrom Point (f) to Point (i) will be equal to the negative of above Item(a) result:

{Work Done ON Fluid For (f) --->(i)} = 1.20e+7 Joules

Work Done ONFluid For (i) ---> (f)} = -{Area Under Curve} =

\(\displaystyle=-{\left({6.0}{e}+{6}\right)}\cdot{\left({1}\right)}\pm{\left({\frac{{{1}}}{{{2}}}}\right)}\cdot{\left({6.0}{e}+{6}+{2.0}{e}+{6}\right)}\cdot{\left({1}\right)}\pm{\left({2.0}{e}+{6}\right)}\cdot{\left({1}\right)}\)

\(\displaystyle=-{1.20}{e}+{7}\) Joules

Work done ON fluid during reverse compression along same pathfrom Point (f) to Point (i) will be equal to the negative of above Item(a) result:

{Work Done ON Fluid For (f) --->(i)} = 1.20e+7 Joules