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)


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
  • 1
    $\begingroup$ Yes, if the pressure is constant during the inflating/deflating phase. $\endgroup$ – G. Bergeron Mar 19 '15 at 15:50

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