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

Answer :

(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.

$P\propto \frac{1}{V}$

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:

$1.0atm\times 14mL=P_2\times 8.5mL$

$P_2=1.6atm$

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.

Answer 2

To answer the question above, we assume that the gas is ideal such that,
                         P1V1 = P2V2
Substitute the given values to the equation,
                        1 atm x 14 mL = P2 x 8.5 mL
The value of P2 from the equation is 1.64 atm. 

B. The depth to where the snorkeler is,
                       (1.64 atm) x (10 m / 1 atm) - (1 atm) x (10 m / 1 atm) = 6.4 m
Thus, the snorkeler is 6.4 m deep into the sea.