Answer
$1.41\times10^5\;J$
Work Step by Step
Let $1.4\;m^3$ of water is contained a distance $x$ through the pipe. If $A$ be the cross-sectional area of the pipe.
Therefore, $V=Al=1.4\;m^3$
Force on the water due to pressure difference is: $F=PA$
Therefore, work done on the water by the pressure difference to displace the water by a distance $l$ is given by
$W=Fl$
or, $W=PAl$
or, $W=PV$
or, $W=1\times1.01\times10^5\times1.4\;J$
or, $\boxed{W=1.41\times10^5\;J}$
