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

Answer:

The volume is 1.66 L

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.

Answer 2

Answer:

$V_{2} = 1.753\,L$

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:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$

The final volume is:

$V_{2} = \left(\frac{T_{2}}{T_{1}}\right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{294.15\,K}{305.15\,K}\right)\cdot \left(\frac{1\,atm}{1.32\,atm} \right)\cdot (2.4\,L)$

$V_{2} = 1.753\,L$