Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)` (If water vapour is assumed to be perfect gas).

Question

Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)` (If water vapour is assumed to be perfect gas).

Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)` (If water vapour is assumed to be perfect gas). Find out of the internal energy change. If 1 mole of water is vaporised at 1 bar pressure and `100^(@)C`.
A. `+37.904kJmol^(-1)`
B. `-37.904kJmol^(-1)`
C. `44.096kJmol^(-1)`
D. `-44.096kJmol^(-1)`

Monjur Alahi

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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
`H_(2)O(l)toH_(2)O(g)`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaU=41.00" kJ "mol^(-1)-8.3JK^(-1)mol^(-1)xx373K`
`=41.00-3.096" kJ "mol^(-1)=37.904" kJ "mol^(-1)`

5 views Answered 2023-02-05 15:29