One mole of an ideal gas undergoes a cyclic change ABCD where the (P,V) co-ordinates are A (5,1), B (5,3), C (2,3) and D (2,1). P is atmosphere and V is in litre. Calculate work done along AB, BC, CD and DA and also net work done in the process. Given `1 atmosphere = 1.01 times 10^5 Nm^-2`.

Answer

(i) `1010J (ii) Zero(iii) -404J (iv)Zero ; 606J`

Explanation

(i) work done along AB (figure)
`W_(1)= area ABFEA=EAxxEF`
`=(5xx1.01xx10^(5))xx2xx10^(-3)= 1010J`

(ii) Work done along BC,
`W_(2)= P(dV)= Zero`
(iii) Work done along CD,
`W_(3)= -area CDEFC= -EDxxEF`
`= -(2xx1.01xx10^(5))xx2xx10^(-3)= -404J`
(iv) work done along DA, `W_(4)=P(dV)= Zero`
Net work done in process
`W=W_(1)+W_(2)+W_(3)+W(4)`
`=1010+0-404+0= 606J`