The latent heat of vapourization of a liquid at $$500\,K$$ and $$1\,atm$$ pressure is $$10.0\,kcal/mol.$$ What will be the change in internal energy $$\left( {\Delta U} \right)$$ of $$3\,moles$$ of liquid at the same temperature
A.
$$13.0\,kcal/mol$$
B.
$$ - 13.0\,kcal/mol$$
C.
$$27.0\,kcal$$
D.
$$ - 7.0\,kcal/mol$$
Answer :
$$27.0\,kcal$$
Solution :
$$\eqalign{
& 3{H_2}O\left( l \right) \to 3{H_2}O\left( g \right); \cr
& \Delta n = 3,\,\Delta E = \Delta H - \Delta nRT \cr
& = 30 - 3 \times \frac{2}{{1000}} \times 500 \cr
& = 27\,kcal \cr} $$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$