Question

Latent heat of vaporization of a liquid at $500$K and $1$atm pressure is $10.00$ kcal/mol. What will be the change in internal energy ($\Delta$U) for $3$ mol of liquid at the same temperature?

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

Answer 1

The correct option is C $27.0$ $kcal$

Given:
n=3 mol, T=500 K, p=1 atm, $\Delta H_{vap}=10kcal/mol$
For 3 mol, $\Delta H_{vap}=30kcal$
$3H_{2}O\left(l\right)\to 3H_{2}O\left(g\right)$

$\Delta ng=3-0=3$

$\Delta E=\Delta H_{vap}-\Delta ngRT$

$=3\times 10-3\times 500\times 0.002$

$=27kcal$