Standard enthaply of vaporisation `Delta_(vap)H^(theta)` for water at `100^(@)C` is `40.66 kJ mol^(-1)`. The internal energy of vapourisation of water at `100^(@)C` (in `kJ mol^(-1)`) is :

Options

A. `+ 40.66`
B. `+ 37.56`
C. `- 43.76`
D. `+ 43.76`

Answer

B

Explanation

`Delta H = Delta U + Delta n_(g) RT`
`H_(2)O(l) rarr H_(2)O(g)`
`Delta n_(g) = 1, Delta H = 40.66 xx 10^(3) Jmol^(-1)`,
T = 373K
`40660 = Delta U + 1 xx 8.314 xx 373`
`40660 = Delta U + 3101.12`
`Delta U = 40660 - 3101.12`
`= 37.558 Jmol^(-1)`
`= 37. 56 kJmol^(-1)`