Im taking my chemistry final right

a snorkeler takes a breath of air so that his lungs are 1,159.49 ml in volume when the pressure is 764.96 mm of hg. if he dives to a depth where the pressure is 1.16 atm, what volume, in ml, will his lungs have? assume the temperature of the water is constant.

When the diver dives to a depth where the pressure is 1.16 atm, he has 1006.1 mL of air in his lungs.

Explanation:

Step 1: Data given

Volume in his lungs = 1159.49 mL

The pressure = 764.96 mmHg = 764.96 / 760 = 1.00652646 atm

When he dives the pressure changes to 1.16 atm.

Step 2: Calculate final volule

(P1*V1)/T1 = (P2*V2)/T2

 ⇒ Since the temperature will be constant we can simplify as:

P1*V1 = P2*V2

⇒ with P1 = The initial pressure 1.00652646 atm

 ⇒ with V1 = The initial volume = 1159.49 mL

 ⇒ with P2 = The pressure after diving = 1.16 atm

 ⇒ with V2 = The volume of air in his lungs after diving

V2 = (P1*V1)/P2

V2 =(1.00652646 * 1159.49)/1.16

V2 = 1006.1 mL

When the diver dives to a depth where the pressure is 1.16 atm, he has 1006.1 mL of air in his lungs.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.