1.

One mole of water at `100^(@)C` is converted into steam at a constant pressure of 1 atm. The change in entropy is [heat of vaporisation of water at `100^(@)C = 540 cal//gm`]A. 8.74B. 18.76C. 24.06D. 26.06

Answer» Correct Answer - 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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