An insulated container contains 1 mole of a liquid, molar volume 100 mL, at 1 bar. When liquid is steeply pressed to 100 bar, volume decreases to 99 mL. Find the value of $\frac{\mathrm{\Delta H}+\mathrm{\Delta U}}{1000}-2$

For adiabatic process, W = P (V₂  -  V₁)

$P_1=1\mathrm{bar},P_2=100\mathrm{bar},V_1=100\mathrm{mL},V_2=99\mathrm{mL}$

For adiabatic process, $q=0 ;\mathrm{\Delta U}=\mathrm{W}$

$\begin{array}{l}\mathrm{\Delta U}=\mathrm{q}+\mathrm{W}\\ =\mathrm{q}-\mathrm{P}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\text{ since }\mathrm{W}=-\mathrm{P}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\\ =0-100(99-100)=100\text{ bar mL }\\ \mathrm{\Delta H}=\mathrm{\Delta U}+\mathrm{\Delta }\left(\mathrm{PV})=\mathrm{\Delta U}+\left({\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{P}}_1{\mathrm{V}}_1\right)\right.\\ =100+\left[\right(100\times 99)-(1\times 100\left)\right]\\ =100+(9900-100)=9900\text{ bar }\mathrm{mL}\\ \mathrm{\Delta U}+\mathrm{\Delta H}=100+9900=10000\\ \therefore \text{ the value of }\frac{\mathrm{\Delta U}+\mathrm{\Delta H}}{1000}-2=\frac{10000}{1000}-2=8\end{array}$