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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 cannot be calculated 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/V_i)=5 "In"373/298+2"In"10/1`
`DeltaH=nC_pDeltaT=7(75)=525` cal
`DeltaE=nC_vDeltaT=5(75)=375` cal
`DeltaG=DeltaH-Delta(TS)`
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