Physics, 21.07.2019 11:30 LuluMathLover101

Adiver 40 m deep in 10∘c fresh water exhales a 1.0-cm-diameter bubble. what is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘c? assume that the air bubble is always in thermal equilibrium with the surrounding water. express your answer to two significant figures and include the appropriate units.

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Answer from: quoia89

Applying gas law:
PV = nRT => nR = PV/T

nR = Constant at all conditions and thus;
PV/T = Constant.
Therefore.
P1V1/T1 = P2V2/T2

Where P = Pressure, V = Volume, T = Temperature in Kelvin

State 1: At 40 m deep
P1 = Atmospheric pressure + water pressure = 101325 + rho*g*h = 101325+1000*9.81*40 = 493725 Pa
V1 = Volume of the bubble = 4/3*πd^3/64 = π*1^2/4 = 0.0654 cm^3
T1 = 10+273.15 = 283.15 K

State 2: At the surface of the lake
P2 = Atmospheric pressure = 101325 Pa
T2 = 20+273.15 = 293.15 K
V2 = ?

Solving for V2
V2 = (T2*P1V1)/(T1*P2) = (293.15*493725*0.0654)/(283.15*101325) = 0.3302 cm^3

But,
V2 = 4/3*πD^3/64 => D = cube root (3/4*V2*64) = 2.512 cm

The diameter of the bubbles as it reaches the surface is 2.5 cm rounded to two significant figures.

Answer from: itaheart101

The diameter of the air bubble when it reaches the surface is about 1.7 cm

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Further explanation

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

$\large {\boxed {P = F \div A} }$

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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In this problem , we will use Ideal Gas Law as follows:

Given:

initial diameter of bubble = d₁ = 1.0 cm

initial depth of the diver = h = 40 m

initial temperature = T₁ = 10 + 273 = 283 K

atmospheric pressure = Po = 1.0 atm = 10⁵ Pa

final temperature = T₂ = 20 + 273 = 293 K

density of water = ρ = 1000 kg/m³

Unknown:

final diameter of bubble = d₂ = ?

Solution:

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

$\frac{P_1 (\frac{1}{6} \pi (d_1)^3)}{T_1} = \frac{P_2(\frac{1}{6} \pi (d_2)^3)}{T_2}$

$\frac{P_1 (d_1)^3}{T_1} = \frac{P_2 (d_2)^3}{T_2}$

$\frac{(P_o + \rho g h ) (d_1)^3}{T_1} = \frac{P_o (d_2)^3}{T_2}$

$\frac{(10^5 + 1000(9.8)(40) ) (1.0)^3}{283} = \frac{10^5(d_2)^3}{293}$

$\frac{(492000 ) (1.0)^3}{283} = \frac{10^5(d_2)^3}{293}$

$d_2 \approx 1.7 \texttt{ cm}$

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Learn moreMinimum Coefficient of Static Friction : The Pressure In A Sealed Plastic Container : Effect of Earth’s Gravity on Objects :

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

Adiver 40 m deep in 10∘c fresh water exhales a 1.0-cm-diameter bubble. what is the bubble's diameter

Answer from: Quest

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step-by-step explanation:

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30 points! and explain and i will give brainliest.

Answer from: Quest

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

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Adiver 40 m deep in 10∘c fresh water exhales a 1.0-cm-diameter bubble. what is the bubble's diameter...