\(W = -Pext\,\,\,\, \Delta V = -1(1.5 - 1.2)\)
\(W = -0.3\, Lit.-atm \,\,\,\,\,\, \,W = -7.26 \,Cal\)
So, \(\Delta U = q + W = 992.74 \,Cal → \Delta U = 0.99274 \,K\, Cal\)
$(i)$ $NH_3$ $_{(g)} + aq$ $\rightarrow$ $NH_3$ $_{(aq)}$, $\Delta H$ $= -8.4 \,Kcal.$
$(ii)$ $HCl_{(g)} + aq$ $\rightarrow$ $HCl{(aq)}$, $\Delta H =$ ${-1}7.3\, Kcal.$
$(iii)$ $NH_3$ $_{(aq)} + HCl_{(aq)}$ $\rightarrow$ $NH_4Cl $ $_{(aq)}$, $\Delta H = -12.5\, Kcal$.
$(iv)$ $NH_4Cl$ $_{(s)} + aq$ $\rightarrow$ $NH_4Cl$ $_{(aq)}$, $\Delta H = +3.9 \,Kcal.$