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I am introducing myself to fluids through some physics textbooks. Most of them start with Pascal's principle.:

If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount

Then there's the example of the hydraulic lift that makes use of Pascal's principle.

What I don't get is why the fluid used in the example is always a liquid (and not a gas for example)? I know that liquids are incompressible but what happens if you use a gas to lift the car? Does part of the input force (or pressure) get wasted in increasing the temperature of the gas and therefore making it less effective? How much would the car get lifted if at all?

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    $\begingroup$ With a gas you need to put in work to compress the gas to the required pressure before you can start to lift the car. Since liquids are almost incompressible very little work is needed to raise the liquid to the required pressure. $\endgroup$ – John Rennie Feb 17 '14 at 18:00
  • $\begingroup$ @John: Right. My simplistic analogy is pushing against a spring. You can't exert a certain force against it without also compressing it. $\endgroup$ – Mike Dunlavey Feb 18 '14 at 18:14
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I think that you're basically right when you make the following suggestion:

Does part of the input force (or pressure) get wasted in increasing the temperature of the gas and therefore making it less effective?

Recall from the First Law of thermodynamics, \begin{align} \Delta E = Q-W \end{align} where $\Delta E$ is the change in internal energy of the fluid, $W$ is the work done by the fluid, and $Q$ is the heat transferred to the fluid during the lifting process. Let's assume that the fluid is thermally insulated so that $Q=0$, then we have \begin{align} \Delta E = -W \end{align} The $W$ term will then be a combination of $W_1$ and $W_2$ where $W_1$ is the magnitude of the work you do to compress the fluid, and $W_2$ is the magnitude of the work done by the fluid in raising the object. In fact, we have \begin{align} W = W_2-W_1 \end{align} so that \begin{align} \Delta E = W_1 - W_2. \end{align} Now, for an incompressible liquid, no work needs to be done to compress the fluid, so we have $W_1=W_2$, and therefore $\Delta E = 0$; the internal energy of the fluid remains the same. However, as suggested by John Rennie, in the case of a gas, some work goes into compressing the gas, and we have $W_2 < W_1$, so we have \begin{align} \Delta E >0 \end{align} But the internal energy of a (ideal) gas is proportional to its temperature, so this means that its temperature increases a bit during the lifting process.

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