Standard enthalpy of vaporisation`DeltaV_(vap).H^(Theta)` for water at `100^(@)C`is`40.66kJmol^(-1)`.The internal energy of Vaporization of water at` 100^(@)C("in kJ mol"^(-1))`is

Options

A. 37.56
B. `-43.736`
C. `+43.76`
D. `+40.56`

Answer

A

Explanation

`Delta_("vap")H^@ = 40.66 kJ mol^(-1)`
`T = 100 + 273 =373 K. Delta U = Delta H - Delta n_g RT`
`Delta n_g` = number of gaseous mole of products number of gaseous moles of reactants
`H_2O(l) =H_2O_((g)), therefore n_g =1 -0=1 `
`Delta U = Delta H- Delta n RT`
`Delta U = (40. 66 xx 10^3)-(8.314 xx 373)`
`=37.56 kJ mol^(-1)`