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Mathematics 16 Online
OpenStudy (anonymous):

WILL MEDAL AND FAN!!!!! PLEASE HELP! A snorkeler takes a syringe filled with 14mL of air from the surface, where the pressure is 1.0 atm, to an unknown depth. The volume of the air in the syringe at this depth is 8.5mL . a) What is the pressure at this depth? b) If the pressure increases by 1 atm for every additional 10 m of depth, how deep is the snorkeler? The answer to part a) is 1.6 atm. How do you solve part b) ?

OpenStudy (anonymous):

@dumbcow

OpenStudy (anonymous):

@cz1

OpenStudy (anonymous):

please help

OpenStudy (dumbcow):

divide the rate by 10 ---> Increase of .1 atm for every 1 m in depth it increased by .65 atm so depth = 6.5

OpenStudy (anonymous):

@dumbcow How did it increase by .65 atm

OpenStudy (dumbcow):

from 1 atm to 1.65 atm i see you rounded it to 1.6

OpenStudy (paxpolaris):

part a) PV=nRT so in this case, Boyle's law applies\[P \propto \frac 1V\]

OpenStudy (anonymous):

x = .607

OpenStudy (perl):

now add that to 1

OpenStudy (anonymous):

x = 1.607

OpenStudy (perl):

woops, let me do part a) again

OpenStudy (paxpolaris):

for part b) we have a linear relationship.... so, (change in depth) / ( change in pressure) is constant

OpenStudy (perl):

PV = nRT , PV = k, since n, R, and T are constant in the problem (we can assume that) so we have P1 *V1 = k P2 *V2 = k substitute P1 * V1 = P2 * V2

OpenStudy (anonymous):

For part a), I converted mL to L, and got 1.647atm. I'm still stuck with part b)

OpenStudy (perl):

you dont need to

OpenStudy (perl):

(1 atm)*(14 mL) = (x atm) * (8.5 mL) , since the units cancel

OpenStudy (perl):

yes i got 1.647 as well

OpenStudy (perl):

its a good idea though to change to Liters , in general

OpenStudy (anonymous):

ok, thanks. could you please explain how to do part b

OpenStudy (paxpolaris):

Change in Depth varies directly as change in Pressure

OpenStudy (perl):

you can make an equation we have P = 1 atm , Depth = 0 P = 2 atm , Depth = 10 meters

OpenStudy (dumbcow):

@koook the change in pressure is .647 right for every .1 the depth goes down 1 meter thus depth = 6.47 m

OpenStudy (perl):

Depth = 10*P -10

OpenStudy (perl):

now plug in pressure from part a)

OpenStudy (perl):

Depth = 10(1.647) - 10

OpenStudy (anonymous):

THANK YOU SO MUCH @perl and @dumbcow and @PaxPolaris

OpenStudy (dumbcow):

:)

OpenStudy (perl):

good job everyone

OpenStudy (anonymous):

case closed :)

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