Pressure in a Lake: Find Depth for 4.5 atm

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To find the depth in a freshwater lake where the water pressure is 4.5 atm, the equation P = a + hgp is used, where P is the total pressure, a is atmospheric pressure, h is depth, g is acceleration due to gravity, and p is the density of water. The calculation shows that h = (4.5x10^5 - 1x10^5) / (9.8 * 1000) results in approximately 35.71 meters. It's crucial to convert all pressures to pascals and ensure the density of freshwater is correctly used in kg/m³. The discussion highlights that 10 meters of water roughly equals 1 atm, suggesting that 35 meters would yield about 4.5 atm total pressure. Proper unit conversion is essential for accurate results in such calculations.
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Homework Statement


At the surface of a freshwater lake the air pressure is 1 atm. At what depth under water in the lake is the water pressure 4.5 atm?

Homework Equations


P=a+hgp
h=(P-a)/(g*p)

The Attempt at a Solution


h=(4.5x105 - 1x105)/(9.8*1000)=35.71
I also did the equation with 9,000 Pa converted to atm, which I think it what I'm supposed to do but I still get the incorrect answer.
 
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Convert everything to pascals before using the equations. Also, be sure that the density you have for freshwater is in kg/m3. I'm not seeing anything wrong in your method, so I'm assuming the problem is in the units.
 
"h=(4.5x105 - 1x105)/(9.8*1000)=35.71" Appears to be correct. If I am not mistaken, 10 m of water is roughly = to 1 atm. But, I am not 100% sure off the top of my head. If that's the case, it would make sense that 35 m of water would be about 3.5 atm plus the 1 atm from the atmosphere = 4.5 atm.

If you know, what is the answer supposed to be?
 
Gear300 said:
Convert everything to pascals before using the equations. Also, be sure that the density you have for freshwater is in kg/m3. I'm not seeing anything wrong in your method, so I'm assuming the problem is in the units.

Thanks, I just needed to convert the top to pascals, I thought I only had to do it to the bottom.
 
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