Question

The latent heat of vaporisation of a liquid at $500K$ and a pressure is $10.0kcal/mol$. What will be the change in internal energy ($\Delta E$) of $3$ moles of liquid at the same temperature?

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

Answer 1

Answer: (C)

$27.0kcal$

Solution

latent heat of vaporization, $\Delta H=13kcal/mol$

T = 500 k P = 1 atom (as std case )

For three moles, $\Delta H=3\times 10=30Kcalmo{l}^{-1}$

Now,

$H=U+PV\Rightarrow U=HPV$

$\Rightarrow \Delta U=\Delta H-\Delta \left(PV\right)$

$\Rightarrow \Delta U=\Delta H-\Delta \left(P\frac{nRT}{P}\right)[\because PV=n,RT]$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

Reaction : $3A\left(L\right)\to 3A\left(g\right)$ $\therefore \Delta ng=3-0=3$

Hence, Change in internal energy, $\Delta U=\Delta H-\Delta ngRT$

$\Rightarrow \Delta =30-3\times 0.002\times 500$

$\Rightarrow \Delta U=30-3=27kcal$

Correct Ans :- (C)