Respuesta :
It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.
By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then
[tex]p_{1}[/tex] =patm+[tex]pH_{2} O[/tex]
[tex]p_{1}[/tex] = 101325+ρ[tex]gh_{1}[/tex]
It is given that height is 125ft. Put the value of h in above formula:
h1 =125ft=38.1m
ρ=1.04g/mL=1040kg/[tex]m^{3}[/tex]
g=9.81
[tex]p_{1}[/tex] =101325Pa+388711.44
[tex]p_{1}[/tex] =490036.44Pa
[tex]p_{2}[/tex] =p atm =101325Pa
It is known that volume and pressure can be expressed as:
V*P=const.
where, V is volume and P is pressure.
Now,
[tex]V_{1} *p_{1}[/tex] =[tex]V_{2} *p_{2}[/tex]
[tex]V_{2} /V_{1}[/tex]=[tex]p_{2} /p_{1}[/tex]
[tex]V_{2} /V_{1}[/tex] =490036.44/101325
[tex]V_{2} /V_{1}[/tex]=4.84
Assume constant temperature
d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.
now [tex]p_{1}[/tex] =p atm+[tex]pH_{2} O[/tex] =490036.44Pa
V*p=const
where, V is volume and P is pressure.
Now,
[tex]V_{1} *p_{1}[/tex] =[tex]V_{2} *p_{2}[/tex]
[tex]V_{2} /V_{1}[/tex]=[tex]p_{2} /p_{1}[/tex]
[tex]V_{2} /V_{1}[/tex] =490036.44/X
[tex]p_{2}[/tex] =490036.44pa/(V2/V1) =326690.96Pa
[tex]p_{2}[/tex] =patm +p[tex]H_{2} O[/tex]
[tex]p_{2}[/tex] =101325Pa+ρ[tex]gh_{2}[/tex]
326690.96Pa=101325Pa+ρ[tex]gh_{2}[/tex]
ρgh1 =151987.5-101325=225365.96Pa
ρ=1,04g/mL=1040kg/m3
g=9.81
[tex]h_{2}[/tex] =225365.96/ρ∗g
[tex]h_{2}[/tex] =225365.96 / 1040∗9.81
[tex]h_{2}[/tex] =22.09m= 72.47ft
ΔH=[tex]H_{1} -H_{2}[/tex]
=125-72.47
=52.53ft
So she can safely ascend up to 52.53 ft without Breathing out
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