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 $\Delta U$.
$(2004,2 M)$
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Solution:
$ \Delta U=q+W $
For adiabatic process, $q=0$, hence $\Delta U=W$
$ W=-p(\Delta V)=-p\left(V_2-V_1\right) $
$ \Rightarrow \Delta U=-100(99-100)=100 \text { bar } \mathrm{mL} $
$ \Delta H=\Delta U+\Delta(p V) $
$ \text { where, } \quad \Delta p V=p_2 V_2-p_1 V_1 $
BETA
