109k views
0 votes
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?

2 Answers

6 votes

Answer:


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

User Kartik Prasad
by
7.8k points
4 votes

Answer:

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.

User Oscaroscar
by
7.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.