Question

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

• A. -13.0 kcal
• B. -27.0 kcal
• C. 27.0 kcal
• D. 13.0 kcal

Answer 1

Answer: (C)

The relation between $\Delta H$ and $\Delta E$ is given by
$\Delta H=\Delta E+\Delta n_{g}RT$
where $\Delta n_g=$ change in the number of gaseous moles.
Vaporisation of $3$ moles of water is
$3H_{2}O(l)⟶3H_{2}O(g)$
$\Delta n_g=3-0=3$
Latent heat of vaporisation $=10$ kcal / mol
So, heat change for 3 moles of water to vapours
$=3\times 10=30$ kcal
Here,
$T=500K$
$R=0.002kcalmo{l}^{-1}{K}^{-1}$
$\Delta H=30$ kcal
Put these value in the above relation
$30=\Delta E+3\times 0.002\times 500$
$30=\Delta E+3$
$\Delta E=30-3=27$ kcal