Pressure of air in cylinder under water

[frostchaos123]

Homework Statement

Referring to the attachment, a cylinder of 2.5m filled with air is submerged to a dept of 82.3m into the sea such that sea water in cylinder is now x m. Water at bottom is 277.15K.

The Attempt at a Solution

Using ideal gas law PV=nRT of air,

the answer gives (1atm + 82.3*rho*g) * (2.5 - x) * Area of cylinder = nR(277.15)

However i don't understand why the pressure of the gas is 1 atm + 82.3*rho*g. Since the pressure of gas is dependant on height of water, shouldn't it be 1 atm + (82.3-x) * rho*g instead?

Or another reasoning is shouldn't it be like pressure of air + pressure of water trapped in cylinder = pressure at the bottom of the sea?

 

[tiny-tim]

hi frostchaos123!

(have a rho: ρ )

However i don't understand why the pressure of the gas is 1 atm + 82.3*rho*g. Since the pressure of gas is dependant on height of water, shouldn't it be 1 atm + (82.3-x) * rho*g instead?

yes, i think you're right …

P is the pressure on the volume of air, which is the pressure at the air-water surface, which is at height 82.3-x

(x will be very small compared with 82.3, but I'm not sure it's small enough to be negligible)

 

[frostchaos123]

Thanks for the help :)

 

[donutz610]

pressure is based on both volume and temperature of water, that might help u understand the problem a little better