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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 |
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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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