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An ideal gas undergoes an expansion from a state with temperature `T_(1)` and volume `V_(1)` through three different polytropic processes `A, B` and `C` as shown in the `P - V` diagram. If `|Delta E_(A)|, |Delta E_(B)|` and `|Delta E_(C )|` be the magnitude of changes in internal energy along the three paths respectively, then :
image
A. `|Delta E_(A)|lt|Delta E_(B)|lt|Delta E_(C )|` if temperature in every process decreases
B. `|Delta E_(A)|gt|Delta E_(B)|gt|Delta E_(C )|` if temperature in every process decreases
C. `|Delta E_(A)|gt|Delta E_(B)|gt|Delta E_(C )|` if temperature in every process increases
D. `|Delta E_(B)|lt|Delta E_(A)|lt|Delta E_(C )|` if temperature in every process increases

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Best answer
Correct Answer - A::C
Initial state is same for all the three processes (say initial internal energy `=E_(0))`.
In the final state, `V_(A)=V_(B)=V_(C)`
and `P_(A) gt P_(B) gt P_(C)`
`P_(A)V_(A) gt P_(B)V_(B) gt P_(C)V_(C)`
` E_(A) gt E_(B) gt E_(C)`
If `T_(1) lt T_(2)`, then `E_(0) gt E_(f)` for all the three processes and hence `(E_(0)-E_(A)) lt (E_(0)-E_(B)) lt (E_(0)-E_(C))`
`|DeltaE_(A)|lt|DeltaE_(B)|lt|DeltaE_(C)|`
If `T_(1)lt T_(2)`, then ` E_(0) lt E_(f)` for all three processes and hence `(E_(A)-E_(0)) gt (E_(B)-E_(0)) gt (E_(C)-E_(0))`
`|DeltaE_(A)|gt|DeltaE_(B)|gt|DeltaE_(C)|`

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