Question

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?

Answer 1

(a) The pressure at this depth is, 1.6 atm

(b) The depth of the snorkeler is, 6 meter.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

`P\propto (1)/(V)`

or,

`P_1V_1=P_2V_2`

where,

`P_1` = initial pressure = 1.0 atm

`P_2` = final pressure = ?

`V_1` = initial volume = 14 mL

`V_2` = final volume = 8.5 mL

Now put all the given values in the above equation, we get:

`1.0atm* 14mL=P_2* 8.5mL`

`P_2=1.6atm`

The pressure at this depth is, 1.6 atm

Now we have to calculate the depth of the snorkeler.

Let the depth of the snorkeler be 'x'.

As per question,

1.6 atm - 1 atm = x meter

0.6 atm = x meter

and,

As, 1 atm = 10 meter

So, 0.6 atm = 6 meter

The depth of the snorkeler is, 6 meter.