One mole of an ideal gas is taken through the cyclic process ABCDA, as shown in (figure). Using the graph, calculate
(i) Work done in the processes `ArarrB, BrarrC, CrarrD and DrarrA`
(ii) Work done in complete cycle ABCDA
(iii) Heat rejected by the gas in one complete cycle.

Explanation

(i) In process A to B, volume is constant
`:. dW_(AB)= P.dV= Zero`
In process BC
`dW_(BC)= area BEFC =BExxBC`
`=(5xx1.013xx10^(5)N//m^(2)xx50)J`
`=2.53xx10^(7)J`
In process CD, volume remains constant
`:. dW_(CD)= P.dV=Zero`
In process DA
`dW_(DA)= area AEFD= AExxEF`
`=(10xx1.013xx10^(5))xx(-50)`
`= -5.06xx10^(7)J`.
(ii) Net work done in complete cycle ABCDA
`dW=dW_(AB)+dW_(BC)+dW_(CD)+dW_(DA)`
`= 0+2.53xx10^(7)+0-5.06xxc10^(7)`
`= -2.53xx10^(7)J`
(iii) As the process is cyclic, `dU=0`.
From first law of thermodynamics
`dQ=dU+dW`
`= 0-2.53xx10^(7)= -2.53xx10^(7)J`
Negative sign shows that this much heat is rejected in one complete cycle.