Thermodynamics and Thermochemistry 1 Question 63
66. $100 \mathrm{~mL}$ 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 $1 \mathrm{~mL}$ 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 $\Delta U=W$
$$ \begin{aligned} W & =-p(\Delta V)=-p\left(V_{2}-V_{1}\right) \ \Rightarrow \quad \Delta U & =-100(99-100)=100 \text { bar } \mathrm{mL} \ \Delta H & =\Delta U+\Delta(p V) \end{aligned} $$
where, $\Delta p V=p_{2} V_{2}-p_{1} V_{1}$
