Question

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

• A. $+13.0Kcal$
• B. $-13.0Kcal$
• C. $-27.0Kcal$
• D. $+27.0Kcal$

Answer 1

Answer: (D)

$+27.0Kcal$

$\Delta H=10Kcal/mol$

$3$ Liquid $⟶3$ Gas

$\Delta n_g=3\times 1=3$, Here $n=3$

Now,

$\Delta U=n\Delta H-\Delta ngRT$

$\Delta U=(10\times 3)-(3\times 2\times 500\times {10}^{-3})$

$\Delta U=(30-3)Kcal$

$\Delta U=27.0Kcal$