# Absolute pressure of air inside a meat baster

1. Dec 28, 2011

### phosgene

1. The problem statement, all variables and given/known data

A meat baster consists of a flexible bulb attached to a plastic tube as shown in the diagram. When the open end of the tube is immersed in the basting sauce and the bulb is squeezed then released, sauce rises in the tube so that it can be squirted over the meat. Suppose sauce rises in the tube to a height h of 0.12m. Assume that the density of the sauce is 1050 kg/m^3

(c) Calculate the absolute pressure of the air in the bulb when atmospheric pressure is 1.010 x 10^5 Pa

2. Relevant equations

Bernoulli's equation to calculate pressure difference between bottom and top points of the height the sauce rises.

3. The attempt at a solution

I calculated the pressure difference betewen the start and end point of the measured height:

P = pg(y2 - y1)
P = (1050 kg/m^3)(9.8 m/s^2)(0.12 m)
P = 100.8 Pa

I'm not really sure how to answer the question, but here's my attempt:

If the sauce isn't rising anymore, then the forces acting on it should be in equilibrium, cancelling each other out. The liquid's force will be directed downward, out of the baster, the atmospheric force will be directed upward, into the baster, and the force of the air inside the baster will be directed downward (pushing the sauce out of the baster). So the atmospheric pressure should be the same magnitude as the other two pressures.

If I do the calculation, I get the absolute pressure of the air in the bulb being 1.010 x 10^5 Pa - 100.8 Pa

But I'm not really sure.

2. Dec 28, 2011

### Staff: Mentor

You'll want to recheck that calculation.
It would appear that your logic is okay, it's just your value for the pressure due to the column of liquid that is suspect.

3. Dec 28, 2011

### Redbelly98

Staff Emeritus
phosgene, welcome to Physics Forums.

Your method is correct. Since as you said you are calculating a pressure difference, the air pressure in the baster differs from atmospheric pressure by the amount you calculate, and in fact is less than atmospheric in order to draw the liquid upward into the tube.

The only problem I see is that you probably should redo this calculation:
It seems to me there is an arithmetic error in what you did.

EDIT: gneill beat me to it.

4. Dec 28, 2011

### phosgene

You're both right, I messed up the calculation. I did it again and got 1234.8 Pa. Thanks for the replies :)

5. Dec 28, 2011

### Redbelly98

Staff Emeritus
You're welcome.

By the way, when I looked at the numbers, it was apparent that the "9.8" and "0.12" sort of cancel each other out, leaving a result that must be around 1000. It's nice when you can check computations in this way and catch those errors (which we all make from time to time).