Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `3` mol of liquid at same temperature?

Options

A. 13.0 kcal
B. `-13.0 " kcal "`
C. 27.0 kcal
D. `-27.0 " kcal " `

Answer

C

Explanation

`Delta H = Delta U + Delta n` RT, given `Delta H = 30 kj " mol "^(-1)`
liq `rarr` vapour
`Delta n = n_(g("Product")-n_(g("Ractant")`
`therefore` for 3 moles
`=3-0 =3 therefore Delta H = 3 xx 10 =30 `
`Delta H = Delta U + Delta n RT `
`Delta U = Delta H - Delta nRT = (30 xx 10^3)-( 3 xx 2 xx 500)`
`= (30 xx 10^3)- (3 xx 10^3)=27 xx 10^3 = 27 k cal `