AP Chemistry Chapter 5 help January 1, 2008

If you are working on your Chapter 5 Independent Study, please post questions to this topic.  Kylee asked the following:

I started on the problems and i understand the pressure stuff in the first section but i don’t understand the unit stuff for problems 27 and 29. Its the exact same type of problem so i’m guessing i kinda need to know what I’m doing. I know the equation to use is P(gas)=P(atm) +( or -) h. But since h is in m or cm and P(atm) is in (atm, torr, Pa, etc.) how do you do that?

One thing that will help you is that the unit of Torr is the same as millimeters of mercury, noted as mm Hg.  On the weather reports on TV, they report this in inches of mercury because the average Americans a) don’t know what a millimeter is and b) don’t much care what the atmospheric pressure is.  Anyways, mm Hg is the same as a Torr so when 1 atm equals 760 Torr, this is also 1 atm = 760 mm Hg.  If your height is given in meters or centimeters, just use your metric conversion expertise and move the decimal.  Also, a few other tips:

• Pay attention to if the manometer is closed or open ended.  Closed-end manometers don’t have anything to do with atmospheric pressure and the pressure of the gas is equal to the height difference of the mercury.  For open-ended manometers, if the gas pressure is more than atmospheric, then the mercury will be higher on the atmosphere side so the height difference must be added to the atmospheric pressure.  If the gas pressure is less than atmospheric, the atmosphere will push more on the mercury and the mercury will be higher on the gas side.  This height must be subtracted from the atmospheric pressure.
• More posted as I receive more questions… 

Good luck!  I hope you are all having a great break and are enjoying the snow!