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The volume of a diver's lungs was measured to be 3.6 liters at 1.0 atmosphere of pressure. As the diver descends, assuming that no air is lost, the pressure increases to 3.2 atmospheres. What is the new lung volume? (in liters)The volume of a diver's lungs was measured to be 3.6 liters at 1.0 atmosphere of pressure. As the diver descends, assuming that no air is lost, the pressure increases to 3.2 atmospheres. What is the new lung volume? (in liters)

Respuesta :

The solution to the problem is as follows:

PV=k 

P=pressure, V= volume, k=constant 

V1=3.6L 

P1=1.0(Pascals) 

P2=3.2(Pascals) 

V2=? 

V2= P1(V1)/P2 

V= 1.0(3.6)/(3.2)=1.125L 

Answer= 1.1L 


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skyluke89

Answer:

1.13 L

Explanation:

The problem can be solved by using Boyle's law for ideal gases, which states that the product between pressure and volume of a gas is constant:

[tex]PV=const.[/tex]

which can also be rewritten as

[tex]P_1 V_1 = P_2 V_2[/tex]

in this case, the initial pressure is P1=1.0 atm and the initial volume of the lungs is V1=3.6 L. The final pressure of the lungs is P2=3.2 atm, therefore by substituting these data into the equation, we can find V2, the new volume of the lungs:

[tex]V_2 = \frac{P_1 V_1}{P_2}=\frac{(1.0 atm)(3.6 L)}{(3.2 atm)}=1.13 L[/tex]

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