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

Question

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

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
` 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`

4 views Answered 2023-02-05 15:29