1.

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=540cal/g)

Answer»

8.74
18.76
24.06
26.06

Solution :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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