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# For a given cyclic Asked by raj436830 22nd March 2020, 11:23 AM Let us get the temperatures of each state.

Let us assume at state-1 , we have the state variables Pressure, volume and temperature are Po , Vo and T respectively.

Process 1-2 is at constant volume.  Pressure is proportional to temperature for ideal gas at constant volume.

hence , we have, ( Po / T ) = ( 2 Po / T2 )  .......................(1)

where T2 is temperature at state 2 .    Hence from above equation, T2 = 2T

Process 2-3 is at constant pressure.  Volume is proportional to temperature for ideal gas at constant pressure.

hence , we have, ( Vo / 2T ) = ( 2Vo / T3 )  .......................(2)

where T3 is temperature at state 3 .    Hence from above equation, T3 = 4T

Process 3-4 is at constant volume.  Pressure is proportional to temperature for ideal gas at constant volume.

hence , we have, ( 2Po / 4T ) = ( 2 Po / T4 )  .......................(3)

where T4 is temperature at state 4 .    Hence from above equation, T4 = 2T

Workdone for process 1-2 , W12 = 0 ( constant volume , dV = 0 ) ,

Heat transfer for process 1-2,  Q12 = n Cv ΔT = n (2R)( 2T - T) = 2nRT
{ Cv = no. of degrees of freedom × (R/2) }

Workdone for process 2-3 , W23 = 2Po ( 2Vo - Vo)  = 2 PoVo = 2nRT
{ Cv = [ no. of degrees of freedom × (R/2) + R ] }

Heat transfer for process 2-3,  Q23 = n Cp ΔT = n (3R)( 4T - 2T) = 6nRT

Workdone for process 3-4 , W34 = 0 ( constant volume , dV = 0 ) ,

Heat transfer for process 3-4,  Q34 = n Cv ΔT = n (2R)( 2T - 4T) = - 4nRT

Workdone for process 4-1 , W41 = Po ( Vo - 2Vo) = - PoVo = - nRT

Heat transfer for process4-1,  Q41 = n Cv ΔT = n (3R)( T - 2T) = -3nRT

Net workdone = W12 + W23 + W34 + W41 = 2nRT - nRT = nRT

Heat absorbed by system = Q12 +Q23 = 8nRT

Efficiency η = Net workdone / Absorbed Heat = (nRT) / (8nRT) = 1/8

Hence 1/η = 8
Answered by Expert 22nd March 2020, 2:56 PM
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