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Let's suppose that we want to pump the balloons underwater from the initial volume $V_0$ to the volume $V_1$. The pressure there equals $p_1$ and the atmospheric pressure is $p_0$.

It is claimed that the work needed to pump the balloons consists of work done in order to increase the gas's internal energy and pushing the water apart, i.e.

$$W = c_V n \Delta T + p_1 V_1 - p_0 V_0 $$

But why doesn't the work needed for pushing apart the water equal

\begin{equation} W_w = (p_1 - p_0) (V_1 - V_0) \end{equation}

from the school formula for the work of a decompressing gas, i.e. $W = -p dV$?

(Based on a problem from 62. Polish Physics Olympiad)

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The formula you cite refers to the work done when the change of volume is made at constant pressure, which is not the case here.

Let's say I start with the balloon in the air with volume $V_0$ at pressure $p_0$. The state I want to end up with is the balloon underwater at a depth with pressure $p_1$ and the balloon having volume $V_1$. I cannot directly integrate $pdV$ since the pressure will vary as I blow up the balloon. Intuitively, you can think of the process as such: deflating the balloon at $p_0$ gaining $p_0V_0$ in energy, sinking the 0 volume balloon to the required depth to have a pressure of $p_1$ doing no work since the buoyancy is then null and inflating the balloon at constant pressure $p_1$ using up $p_1 V_1$ energy. Omitting the internal energy, the total work consisting of pushing surrounding fluids (air/water) is $p_1V_1 - p_0V_0$.

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  • $\begingroup$ So we could say that energy connected with the volume of a fluid is the work needed to decompress it from 0 to $V$ at current pressure, thus $E_V = p_kV_k$? $\endgroup$ – marmistrz Mar 19 '15 at 15:27
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    $\begingroup$ Yes, if the pressure is constant during the inflating/deflating phase. $\endgroup$ – G. Bergeron Mar 19 '15 at 15:50

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