If the enthalpy of vaporisation of water is `186.5 J mol^(-1)`, then entropy of its vaporisation will be

Options

A. `4.0 JK^(-1)mol^(-1)`
B. `3.0JK^(-1) mol^(-1)`
C. `1.5JK^(-1) mol^(-1)`
D. `0.5 JK^(-1)mol^(-1)`

Answer

A

Explanation

` Delta _("vap") H = 186.5 J//mol = 0.1865 KJ//mol.`
Temperature T = ` 273 + 100 = 373 K `
` Delta _("vap") H^(@) = 0.1865 KJ//mol`
Entropy changes `Delta S = (Delta_("vap").H^(@))/T`
` T = 273 + 100 = 373 K`
` = (0.1865)/373 = 0.5J//K mol`