Calculating Work Required to Submerge Gas at Constant Temperature and Pressure

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To calculate the work required to submerge 10.0L of air at a constant temperature and pressure to a depth of 10.3m, one must consider the net unbalanced force, which is the difference between the buoyant force and the gravitational force acting on the air. The buoyant force varies with depth, necessitating the use of an integral to account for this change. The formula derived for the net force is Fnet = 981h - 235.44, leading to the integral W = ∫(981h - 235.44) dh from 0 to 10.3m. The calculated work required is approximately 49,612 Joules. This approach effectively addresses the varying forces involved in the submersion process.
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Hi, I have a physics problem that I am not sure how I should be approaching:

Consider a sample consisting of 10.0L of air at absolute pressure 2.00 atm, with density 2.40 kg/m^3. Find the work required to transport it to a depth of 10.3m with its temperature, volume, and pressure remaining constant.

What should I be looking for in this question? My intuition tells me I should be looking for the change in energy as it changes depth. How should I be thinking about this question?

Any helpful hints would be appreciated. Thanks in advance.
 
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Think in terms of forces. Work done is force x distance moved. How much force do you have to exert on the sample to move it to that depth. Also the problem suggests that you will have to do work against the pressure change. (pressure= force/area).
 
Since there is no area mentioned, I am assuming I won't have to deal with pressures.

So, if I calculate the net unbalanced force (Buoyant force - Gravitational = Force Needed to Submerge) and multiply it by the depth, I should get the work required. However, the buoyant force is different at every depth, so how will I overcome this problem?
 
Here is how I attempt to solve it:

B=buoyant force, r(w) = density of water, r(a) = density of air, V=volume of gas, h=depth

Fnet = B - Fg = r(w)gh - r(a)gV = 1000*9.81*h - 2.40*9.81*10 = 981h - 235.44

and.. since the forces vary with depth,

W = integral of (981h -235.44) from 0 to 10.3m,
which gives a resulting work of ~49612 Joules.

Is this correct?
 
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