Question

(0.09
(0,0)
33. Work done for the process shown in the figure is
B(30 kPa, 25 cc)
A(10 kPa, 10 cc)
(1) 15
(3) 4.5 J
(2) 1.5 J
(4) 0.3 J​

Answer 1

(The diagram missing in the question is attached below)

Given :

Initial conditions

      P₁ = 10kPa

      V₁ = 10×10⁻⁶ cubic metres

Final conditions    

      P₂ = 30kPa

      V₂ = 25×10⁻⁶ cubic metres

To Find :

The work done in the process

Solution :

• The work done in the process is given by area under the graph

• Area = Area of the rectangle + Area of the triangle

       $A = \frac{1}{2} \times base\times height+(Area\: of \:the \: rectangle)$

       $A=\frac{1}{2} \times (30-10)(25-10)+(10-0)(25-10)$

       $A=\frac{1}{2} \times (20)(15)+(10)(15)$

       $A=150+150$

       $A=300kPa\times cc$

• By changing the units of work into SI units we get

        $W=0.3\:Joules$

Work done in the process is 0.3J