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One mole of water at 100^@C is converted...

One mole of water at `100^@`C is converted into steam at `100^@`C at a constant pressure of 1 atm. The change in entropy is ____________. (heat of vaporization of water at `100^@`C=540 cal/g)

A

8.74

B

18.76

C

24.06

D

26.06

Text Solution

Verified by Experts

The correct Answer is:
D
The entropy change = `("heat of vaporization")//("temperature")`
Here, heat of vaporisation = 540 cal/gm
`=540xx18 cal mol^(-1)`
Temperature of water = 100 + 273 = 373 K
`therefore` entropy change = `(540xx18)/(373) = 26.06 cal mol^(-1) K^(-1)`.
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