Question

The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calculate the absolute (total) pressure at the bottom of the trench in atmospheres. (1.00 in = 2.54 cm, 1.00 atm = 1.01 × 105 pa)

Answer 1

1110 atm    

Let's start by calculating how many cm deep is 36,000 feet. 

 36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm   

 Now calculate how much a column of water 1 cm square and that tall would mass. 

 1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2   

 We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).  

 It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.   

 1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2 

 11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals   

 Now to convert to atm 

 111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm   

 Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result). 

 1107.274 atm + 1 atm = 1108.274 atm   

 And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.