Question

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? If an expansion factor greater than 1.5 causes lung rupture, how far could she safely ascend from 125 ft without breathing? Assume constant temperature (d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL).

Answer 1

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

$p_{1}$ =patm+$pH_{2} O$

$p_{1}$ = 101325+ρ$gh_{1}$

It is given that height is 125ft. Put the value of h in above formula:

h1  =125ft=38.1m

ρ=1.04g/mL=1040kg/$m^{3}$

g=9.81                                  

$p_{1}$ =101325Pa+388711.44

$p_{1}$   =490036.44‬Pa                

$p_{2}$ =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,              

$V_{1} *p_{1}$ =$V_{2} *p_{2}$

$V_{2} /V_{1}$=$p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/101325

$V_{2} /V_{1}$=4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now $p_{1}$ =p atm​+$pH_{2} O$ =490036.44Pa

V*p=const                              

where, V is volume and P is pressure.

Now,              

$V_{1} *p_{1}$ =$V_{2} *p_{2}$

$V_{2} /V_{1}$=$p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/X

$p_{2}$ =490036.44pa/(V2/V1) =326690.96Pa

$p_{2}$ =patm +p$H_{2} O$

$p_{2}$ =101325Pa+ρ$gh_{2}$

326690.96Pa=101325Pa+ρ$gh_{2}$

ρgh1  =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

$h_{2}$ =225365.96/‬ρ∗g

​$h_{2}$  =225365.96  / ‬1040∗9.81

$h_{2}$  =22.09m= 72.47ft

ΔH=$H_{1} -H_{2}$

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

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