science; AP Chem
posted by Anonymous on .
I don't know how to find the kpa for 35.78 m in this question. I know its a P1V1=P2V2 question, i just need the kpa for 35.78m pls!
Divers know that the pressure exerted by
the water increases about 100 kPa with every
10.2 m of depth. This means that at 10.2 m
below the surface, the pressure is 201 kPa;
at 20.4 m below the surface, the pressure is
301 kPa; and so forth. If the volume of a
balloon is 3.9 L at STP and the temperature
of the water remains the same, what is the
volume 35.78 m below the water’s surface?
Answer in units of L
when i tried to figure it out, i got 451 kpa but i don't think that's right..
If we use the 100 kPa/10.2 m your 451 is very close.
100 x (35.78/10.2) = 452 kPa.
If you figure it by physics
density(g/cc) x depth(m) x gravity(m/s) = 1 g/cc x 35.78 x 9.8 = 350.6 and that plus the 1 atm on top is
350.6 + 101.3 = 450.9.
ooh ok, thank you very much!
this site is seriously a life saver!