What is the value of `Delta H` for a reversible isothermal evaporation of 180g of water at `100^(@)C` ? Latent ehat of evaporation of water `=539.7 cal K^(-1)g^(-1)` ?

Options

A. 5.397kcal
B. 97.146 kcal
C. `(539.7 xx 373)/(1000)kcal`
D. `(10xx373xx539.7)/(1000)kcal`

Answer

B

Explanation

`Delta H` is the heat absorbed at constant pressure. Thus during evaporation (of 180g) of water as the process is done at constant pressure.
`:. Delta H=180 xx 539.7 =97146cal` or `97.146 kcal`