Answer
a) $P_{atm}=96.71kPa$
b) $P_{abs}=138.4kPa$
Work Step by Step
a) Knowing that the $\rho_{water}=1000\frac{kg}{m^3}$ we can calculate the gage pressure at a depth of $9m$
$P_{gage,water}=\rho_{water}*g*h=1000\frac{kg}{m^3}*9.81\frac{m}{s^2}*9m=88.29kPa$
As $P_{abs}=P_{atm}+P_{gage}$
$P_{atm}=P_{abs}-P_{gage}=185kPa-88.29kPa=96.71kPa$
b) $\rho_{liquid}=0.85*1000\frac{kg}{m^3}=850\frac{kg}{m^3}$
$P_{gage,liquid}=\rho_{liquid}*g*h=850\frac{kg}{m^3}*9.81\frac{m}{s^2}*5m=41.69kPa$
Then:
$P_{abs}=P_{atm}+P_{gage}=96.71kPa+41.69kPa=138.4kPa$
