Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm
Using above equation as:
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
[tex]{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}[/tex]
[tex]{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L[/tex]
[tex]{V_2}=0.27\ L[/tex]
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is [tex]0.27 L[/tex].
Boyle's law, states that at a constant temperature, pressure (p) of a given quantity of gas varies inversely with its volume (v) .
From Boyle's law,
[tex]P1 V1= P2 V2[/tex]
Making V2 subject of formula we have,
[tex]V2= (P1*V1)/P2[/tex]
Where, P1= 1atm, P2= 13.3 atm V1= 3.6L
Then substitute the values, then we have;
[tex]V2= (1*3.6)/13.3 = 0.27L[/tex]
Therefore, the volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is 0.27 L
Learn more about Boyle's law at,;
https://brainly.com/question/24938688
