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An ideal gas undergoes an expansion from...

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 :

A

`|DeltaE_(A)|lt|DeltaE_(B)|lt|DeltaE_(C)|` if temperature in every process decreases

B

`|DeltaE_(A)|gt|DeltaE_(B)|gt|DeltaE_(C)|` if temperature in every process decreases

C

`|DeltaE_(A)|gt|DeltaE_(B)|gt|DeltaE_(C)|` if temperature in every process increases

D

`|DeltaE_(B)|lt|DeltaE_(A)|lt|DeltaE_(C)|` if temperature in every process increases

Text Solution

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The correct Answer is:
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)gtP_(B)gtP_(C)P_(A)V_(A)gtP_(B)V_(B)gtP_(C)V_(C) , E_(A)gtE_(B)gtE_(C)`
if `T_(1)ltT_(2)`, then `E_(0)ltE_(f)` for the three processes and hence
`(E_(0)-E_(A))lt(E_(0)-E_(B))gt(E_(C)-E_(0))`
`|DeltaE_(A)|lt|DeltaE_(B)|lt|DeltaE_(C)|`
If `T_(1) lt T_(2)`, then `E_(0) lt E_(f)` for all the 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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