Thermodynamics and Thermochemistry - Result Question 66
####66. $100 mL$ of a liquid contained in an isolated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by $1 mL$ at this constant pressure. Find the $\Delta H$ and $\Delta U$.
(2004, 2M)
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Solution:
- $\Delta U=q+W$
For adiabatic process, $q=0$, hence $\Delta U=W$
$$ \begin{aligned} W & =-p(\Delta V)=-p\left(V _2-V _1\right) \ \Rightarrow \quad \Delta U & =-100(99-100)=100 \text { bar } mL \ \Delta H & =\Delta U+\Delta(p V) \end{aligned} $$
where, $\Delta p V=p _2 V _2-p _1 V _1$
