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Katena32 [7]
3 years ago
7

A certain flexible weather balloon contains 7.4 L of helium gas. Initially, the balloon is in WP at 8500ft, where the temperatur

e is 20.6oC and the barometric pressure is 577.0 torr. The balloon then is taken to the top of Pike’s Peak at an altitude of 14,100ft, where the pressure is 400 torr and the temperature is 7.5oC. What is the new volume of the balloon at the top of Pikes Peak?
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
3 0
<h2>The new volume of the balloon at the top of Pikes Peak is 10.2 L</h2>

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 577.0 torr

P_2 = final pressure of gas = 400 torr

V_1 = initial volume of gas = 7.4 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 20.6^oC=273+20.6=293.6K

T_2 = final temperature of gas = 7.5^oC=273+7.5=280.5K

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

\frac{577.0\times 7.4}{293.6K}=\frac{400.0\times V_2}{280.5K}

V_2=10.2L

Thus the new volume of the balloon at the top of Pikes Peak is 10.2 L

Learn more about combined gas law

brainly.com/question/12089296

brainly.com/question/4133756

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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
vova2212 [387]

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+ρgh_{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

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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}

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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 +pH_{2} O

p_{2} =101325Pa+ρgh_{2}

326690.96Pa=101325Pa+ρgh_{2}

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ρ=1,04g/mL=1040kg/m3

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h_{2} =225365.96/‬ρ∗g

​h_{2}  =225365.96  / ‬1040∗9.81

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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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