Pressure and variation in pressure due to depth In my class, we were shown the following problem:

Using the equation $P_1$ = $P_0$ + $\displaystyle\rho$gh, where $P_1$ is the pressure at Superman's mouth and $P_0$ is the pressure at the surface of the container he's drinking from, I can find the height that he could pull the fluid up through the vacumn. The first thing my professor said was to assume that the pressure in Superman's mouth can be 0. At this point I'm already lost because I don't understand why the pressure in his mouth would be 0, or what it would mean to have a pressure of 0. Can anybody help me out ?
 A: People who say this is impossible are assuming that Superman drinking from a straw is a static situation. When nothing is moving, the best S. can do is create a perfect vacuum (pressure of 0 psi / Pascals / atmospheres / krypton standard pressure units). HOWEVER...
If he creates a perfect vacuum at the top of a thick straw, the water that gets pushed up will have a certain amount of work done on it - in fact the pressure difference between the top and the bottom of the column is at all times equal to the length of the column. So when the column has reached a height $h$, the work done by the pressure $P$ is $PAh$ But the potential energy of the column of water is $\rho Agh/2$. Keep sucking like that and water could in principle reach a height of 20 meters before the column of water stops moving. So Superman can take a good sip, the let the water settle and repeat.
Another technique to suck water over a greater height (and one that can be maintained indefinitely since it doesn't rely on dynamics) adds bubbles from time to time. pull the straw out of the water and draw a bit of air into it; then put it back and keep sucking. If you make sure there is at least 2 meters' worth of air in the straw you can drink to your heart's content. You will have changed the effective (apparent) density of the water in the straw. You might get a little gassy...
