Answer
$P_{1}=18.12psia$
Work Step by Step
Starting at point $1$ in the natural gas pipeline and assuming that the variation on pressure cause by the aie is negligible:
$P_{1}-\rho_{Hg}*g*h_{Hg}-\rho_{water}*g*h_{water}=P_{atm}$
$P_{1}=P_{atm}+g*(\rho_{Hg}*h_{Hg}+\rho_{water}*h_{water})$
$\rho_{water}=62.4\frac{lbm}{ft^3}$
$\rho_{Hg}=13.6*62.4\frac{lbm}{ft^3}=848.64\frac{lbm}{ft^3}$
$h_{water}=(25in+2in)*(\frac{1ft}{12in})=2.25ft$
$h_{Hg}=(6in)*(\frac{1ft}{12in})=0.5ft$
Substituting:
$P_{1}=14.2psia+(32.2\frac{ft}{s^2})*[(848.64\frac{lbm}{ft^3})*(0.5ft)+(62.4\frac{lbm}{ft^3})*(2.25ft)]*(\frac{1lbf}{32.2\frac{lbm*ft}{s^2}})*(\frac{1ft^2}{144in^2})$
$P_{1}=18.12psia$
