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Question

# One mole of an ideal monatomic gas undergoes the following four reversible processes:Step1 - it is first compressed adiabatically from volume 8.0m3 to 1.0m3.Step2 - then expanded isothermally at temperature T1 to volume 10.0m3.Step3 - then expanded adiabatically to volume 80.0m3.Step4 - then compressed isothermally at temperature T2 to volume 8.0m3.Then T1/T2 is

A
2
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B
4
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C
6
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D
8
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Solution

## The correct option is C 4Given:V1=8m3V2=1m3 V3=10m3V4=80m3To find T1T2=? for ourcase →T2T4=? (as per our diagram)Process: {2→3&4→1}→ Isothermal processT2=T3&T4=T1Process 1→2we know for adiabatic process →T1T2=(V2V1)γ−1⇒T1T2=(18)γ−1 ...(i)3→4 ad. expn.T4T3=(V3V4)γ−1=(18)γ−1 ...(ii)from eqn. (i) & (ii)γ=1.66 T1T2=T4T3⇒T4T2=(18)γ−1⇒T4T2=14( Since T1=T4&[T4T3=(18)γ−1])T2T4=2That is T′1T2=4

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