Question

A diver finds the best corals at a depth of approximately 10.0 m. The diver's lung capacity is 2.4 L. The air temperature is 32.0 °C and the pressure is 1 atm. What is the volume of the diver's lungs at 10.0 m, at a temperature of 21.0°C, and a pressure of 1.39 atm? What gas law is used to solve this problem?

Answer 1

`V_(2) = 1.753\,L`

Step-by-step explanation:

Let assume that air inside lungs behave ideally. The gas laws to be used are Boyle-Mariotte and Charles-Gay-Lussac, whose relationship is:

`(P_(1)\cdot V_(1))/(T_(1)) = (P_(2)\cdot V_(2))/(T_(2))`

The final volume is:

`V_(2) = \left((T_(2))/(T_(1))\right)\cdot \left((P_(1))/(P_(2))\right)\cdot V_(1)`

`V_(2) = \left((294.15\,K)/(305.15\,K)\right)\cdot \left((1\,atm)/(1.32\,atm) \right)\cdot (2.4\,L)`

`V_(2) = 1.753\,L`

Answer 2

The volume is 1.66 L

Step-by-step explanation:

Using General gas equation, we can find the volume of the dive's lungs at 10.0m depth.

P1V1/T1 = P2V2/T2

V2 = P1V1T2 / P2 T1

where; P1 = IATM

P2 = 1.39

V1 = 2.4L

T1 = 32.0 = 32 +273 = 305K

T2 = 21.0 C = 21 +273K = 294K

Equate the values into the equation, we get;

V2 = P1V1T2/ P2T1

V2 = 1 * 2.4 * 294 / 1.39 * 305

V2 = 705.6 / 423.95

V2 = 1.66L

The volume of the diver's lung capacity is 1.66L at the new temperature and pressure.