A snorkeler with a lung capacity of 4.3 L inhales a lungful of air at the surface, where the pressure is 1.0 atm. The snorkeler then descends to a depth of 49 m , where the pressure increases to 5.9 atm. What is the volume of the snorkeler's lungs at this depth? (Assume constant temperature.)

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kobenhavn kobenhavn · Answer 1 · 2020-02-07T04:56:52+01:00

Answer: Volume of the snorkeler's lungs at this depth is 0.73 L

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)

The combined gas equation is,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 1.0 atm

[tex]P_2[/tex] = final pressure of gas = 5.9 atm

[tex]V_1[/tex] = initial volume of gas = 4.3 L

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get:

[tex]{1.0\times 4.3}={5.9\times V_2}[/tex]

[tex]V_2=0.73L[/tex]

Thus the volume of the snorkeler's lungs at this depth is 0.73 L