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!
Question:
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!
To find the kPa at 35.78 m below the water's surface, we can use the given information that the pressure increases by 100 kPa every 10.2 m of depth.
First, we need to determine the number of 10.2 m intervals in 35.78 m:
35.78 m / 10.2 m ≈ 3.51 intervals
Since the pressure increases by 100 kPa for each interval, we can multiply the number of intervals by 100 kPa to find the pressure increase:
3.51 intervals * 100 kPa/interval = 351 kPa
To find the pressure at 35.78 m, we need to add this pressure increase to the pressure at 20.4 m. From the given information, we know that at 20.4 m, the pressure is 301 kPa:
301 kPa + 351 kPa = 652 kPa
Therefore, the pressure at 35.78 m below the water's surface is 652 kPa.
Note that the question is asking for the volume at 35.78 m, not the pressure. However, we have found the pressure value for this depth based on the given information. To find the volume at 35.78 m, we need additional information or equations related to volume.