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.

One mole of an ideal gas is taken through the cyclic process ABCDA, as

1.	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.
Answer» (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.

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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.

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