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One mole of an ideal diatomic gas (O(v)=...

One mole of an ideal diatomic gas `(O_(v)=5 cal )` was transformed from initial `25^(@)C` and 1L , to the state when temperature is `100^(@)C` and volume `10L` . The entropy change of the process can be expressed as `(R=2 ` calories `// mol//K )`

A

`DeltaH=525`

B

`DeltaS=5 "In" (373)/(298) + 2 "In"10`

C

`DeltaE=525`

D

`DeltaG` of the process can not be calculate using given information.

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`DeltaS =nC_(v) "In" ((T_(f))/(T_(i))) + "nR In" ((V_(f))/(T_(i))) =5 "In" (373)/(298) + 2 "In" 10`
`DeltaH = "nC"_(P)DeltaT = n(C_(v)+R) DeltaT =1xx7xx75=525` cal
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