Question

Latent heat of vaporisation of a liquid at $500 \,K$ and $1 \,atm$ pressure is $10.0 \,kcal / mol$. 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
$\Delta H=\Delta E+\Delta n_{g} R T$
where $\Delta n_{g}=$ change in the number of gaseous moles.
Vaporisation of $3$ moles of water is
$3 H _{2} O ( l ) \longrightarrow 3 H _{2} O ( g )$
$\Delta n_{g}=3-0=3$
Latent heat of vaporisation $=10$ kcal / mol
So, heat change for 3 moles of water to vapours
$=3 \times 10=30$ kcal
Here,
$T =500 \,K$
$R =0.002 \,kcal\, mol ^{-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