The pressure at the surface of a lake is `101 kPa` . What depth is the pressure twice atmospheric pressure? What about if the lake was made of mercury?

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The question states that 1 atmosphere of pressure, 101,000 Pa, is already present.  In order to achieve 2 atmospheres of pressure, we need increase our pressure by another 101,000 Pa.  That can be done by descending 10.3 meters down into the lake.  

In order to know what the pressure at a given depth within a fluid is, you can use the formula P = dgh

P = pressure

d = density 

g = gravity (9.8 m/s/s)

h = depth in the fluid (measured in meters)

Water has a density of 1,000 kg/m^3.  From here, we just plug in the numbers and solve for h.  

101,000 = 1,000(9.8)h

101,000 = 9,800h

10.3 meters = h

Use the same formula and switch the density to that of mercury (13,600 kg/m^3) to calculate the depth needed to descend to in mercury in order to achieve an additional atmosphere of pressure.  

101,000 = 13,600(9.8)h

101,000 = 133,280h

.76 meters = h

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The pressure due to a column of height `h` of a liquid of density `rho` is given by its weight per unit area. The pressure as a function of height is then: 

`P(h)=W/A=(mg)/A=((rhoV)g)/A=rho g h `

The pressure at any given depth in a liquid is the sum of the pressure at the surface of the liquid and the pressure due to the liquid at a given depth in the liquid.


`P=P_(at)+rho g h`

Solve for `h` .

`h=(P-P_(at))/(rho g)`

Now let `P=2P_(at)` for twice atmospheric pressure.

`h=(2P_(at)-P_(at))/(rho g)=P_(at)/(rho g)`

Substitute in numerical values.

`h=(101 kPa)/((1.00*10^3 (kg)/m^3)(9.81 m/s^2))`

`h=10.3 m`

Now lets preform this calculation with mercury where `rho_(HG)=13.6*10^3 (kg)/m^3` .

`h=P_(at)/(rho_(HG) g)=(101 kPa)/((13.6*10^3 (kg)/m^3)(9.81 m/s^2))`

`h=75.7 cm`

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