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# physics

Asked by rc1927633 31st May 2022, 11:41 AM
Process AB is constant volume heating process. In this path , TB > TA .

( T is temperature , subscript indicates the state )

Process BC is constant pressure heating process. In this path , TC > TB .

Process CD is constant volume cooling process. In this path , TD < TC .

Process DA is constant pressure cooling process. In this path , TA < TD .

Hence heat is extracted from source only in the paths AB and BC .

Heat extracted in path AB , QAB = n Cv ( TB - TA ) .............................(1)

where n is number of moles of monoatomic gas used in this cycle and
Cv = (3/2)R is constant volume specific heat, R is universal gas constant

Hence we write eqn.(1) as

QAB = n (3/2) R [ TB - TA ] = ( 3/2 ) [ n R TB - n R TA ]

QAB = ( 3/2 ) [ ( 3 Po ) Vo - (2Po ) Vo ] = ( 3/2 ) Po Vo

Heat extracted in path BC , QBC = n Cp ( TC - TB ) .............................(2)

where Cp = (5/2)R is constant pressure specific heat

Hence we write eqn.(2) as

QBC = n (5/2) R [ TC - TB ] = ( 5/2 ) [ n R TC - n R TB ]

QBC = ( 5/2 ) [ ( 3 Po ) 3Vo - (3Po ) Vo ] =  15 Po Vo

Net Heat extracted in the cycle = QAB + QBC  = [ (3/2 + 15 ] Po Vo = 16.5 Po Vo

Answered by Expert 31st May 2022, 12:55 PM
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Tags: heat-energy
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