For an ideal gas graph is shown for three processes. Process 1, 2 and 3 are respectively
A
Isobaric, adiabatic, isochoric
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B
Adiabatic, isobaric, isochoric
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C
Isochoric, adiabatic, isobaric
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D
Isochoric, isobaric, adiabatic
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Solution
The correct option is D Isochoric, isobaric, adiabatic Isochoric process dV=0 W=0 process 1 Isobaric : W=PΔV=nRΔT Adiabatic W=nRΔTγ−1:0<γ−1<1 As workdone in case of adiabatic process is more so process 3 is adiabatic and process 2 is isobaric