Question

Air contained within a piston-cylinder assembly is slowly heated. During this process the pressure first increases linearly with volume where p1 = 100 kPa, p2 = 450 kPa, V1 = 0.4 m3 and V2 = 1.5 m^3. After this initial change the pressure is constant at 450 kPa and the volume expands to 2.1m^3. Determine the total work done in kJ.

Answer 1

Answer:

Total work done ,W = 437.5 kJ

Explanation:

The area under the curve in the PV diagram gives the total work done.

As from the graph, we can find both the area of the trapezium and the area of the rectangle and then sum them to get the total area which is nothing but the total work done under the pressure volume graph when the air in the cylinder is slowly heated.

Therefore,  we get

Area = 1 / 2 ( area of the trapezium ABCFA) + area of the rectangle FCDEF

        = $\frac{1}{2}\times 550\times (1.5-0.4)+450\times (2.1-1.5)$

        = 437.5 kJ