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Two identical bubbles of gas form at the bottom of a lake, then rise to the surface, expanding as they rise (since the pressure is lower nearer the top of the lake). Bubble A rises very quickly, so no heat is exchanged with the water. Bubble B rises slowly, so that it always remains in thermal equilibrium with its surrounding water (which has the same temperature everywhere). Discuss which of the bubbles is larger by the time they hit the surface.

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Kazeemsodikisola

Since the two bubbles are identical

So, they have the same number of molecules

NOTE:

Bubble A undergoes adiabatic compression

For adiabatic compression,

PAVAⁿ = C

Also, Bubble B undergoes isothermal compression

For isothermal compression

PBVB = C

These constant are equal since they are identical and have same number of molecules.

Also, the two bubbles both end at the surface, so they have the same pressure i.e. PA. = PB

So, equating the adiabatic compression constant to isothermal compression constant

Then,

PAVAⁿ = PBVB

Since, PA=PB, then they will cancel out, so we have

VAⁿ = VB

So, VA = n√VB

VA is the nth-root of the VB

n is the adiabatic index and it is always greater than unity

n > 1

So, since n>1

Then, VA < VB

So, bubble B is larger than bubble A.

Also, from ideal gas law

PV=nRT

V = nRT / P

So, the volume depends on the temperature at the surfaces and both bubbles expand and does work on outside environment.

Bubble B is absorbing energy from the environment to maintain constant temperature.

Bubble A did not absorb heat from the lake, so they experience a net loss in internal energy and thus has a temperature lower at the surface

So, bubble B is warmer at the surface than bubble A, then, Bubble B must be larger than Bubble A.

Answer:

The pressure of Bubble B is larger

Explanation:

For bubble A, since no heat is exchanged between the bubble and the water, the process is an adiabatic process.

Let the pressure at the bottom of the lake be = [tex]P_{0}[/tex]

For the adiabatic process: [tex]P_{A} V_{A} ^{k} = P_{0} V_{0} ^{k}[/tex]

Making [tex]V_{A}[/tex] the subject of the formula

[tex]V_{A} = V_{0} (\frac{P_{0} }{P_{A} } )^{1/k}[/tex]...............(1)

For bubble B, since the bubble remains at thermal equilibrium with the water, it is an isothermal process.

For the isothermal process: [tex]P_{B} V_{B} = P_{0} V_{0}[/tex]

Making, [tex]V_{B}[/tex] the subject of the formula

[tex]V_{B} = V_{0} \frac{P_{0} }{P_{B} }[/tex]......................(2)

The bubbles are identical, they possess the same final pressure which is the atmospheric pressure

[tex]P_{A} = P_{B}[/tex] = [tex]P_{atm}[/tex]................(3)

Dividing (2) by (1) and substituting for (3)

[tex]\frac{V_{A} }{V_{B} } = (\frac{P_{0} }{P_{atm} }) ^{\frac{1}{k}-1 }[/tex]

K is always greater than 1 i.e. k > 1

[tex]\frac{V_{A} }{V_{B} } < 1[/tex]

[tex]V_{A} < V_{B}[/tex]

The pressure of Bubble B is larger when the two bubbles hit the surface

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