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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

`3 "In" (298)/(373) + 2 "In" 10`

B

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

C

`7 "In" (373)/(298) + 2 "In" (1)/(10)`

D

`5 "In" (373)/(298) + 2 "In"( 1)/(10)`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaS = nC_(v) "In" ((T_(f))/(T_(i))) + nR "In" ((V_(f))/(V_(i))) = (5)/(2) xx 2 "In" (373)/(298) + 2 "In"10`
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