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If the enthalpy change for the transitio...

If the enthalpy change for the transition of liquid water to steam is `30kJ*mol^(-1)` at `27^(@)C`, the entropy change in `J*mol^(-1)*K^(-1)` for the process would be-

A

10

B

`1.0`

C

0.1

D

100

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaS_("vap")=(DeltaH_("vap"))/(T_(b))=(30xx10^(3)J*mol^(-1))/((273+27)K)`
`=100J*K^(-1)*mol^(-1)`.
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