Atmospheric pressure and water pressure - Boyle's law

[laser]

Homework Statement

See Below

Relevant Equations

PV=constant, P_{fluid} = rho*g*h

I am a bit confused on the marking scheme as attached above.

P1V1 is a constant by Boyle's Law. If the volume increases by a factor of 3, then the pressure decreases by a factor of 3.

This means that the pressure at the top is 1/3 the pressure at the bottom, right? The pressure at the top is just atmospheric pressure (1E5 Pa), so the pressure at the bottom must be (3E5 Pa). 3E5 = Pressure due to water + Pressure due to air, so pressure due to water is just 2E5. So 2E5 = rho*g*h.

Is there a flaw in my understanding? Thanks!

 

[kuruman]

The pressure at the bottom is $p_{\text{bot.}}=p_{\text{atm.}}+\rho g h$ and the temperature (assumed constant) is $T$. What is the initial volume according to the ideal gas law?

 

[laser]

The pressure at the bottom is $p_{\text{bot.}}=p_{\text{atm.}}+\rho g h$ and the temperature (assumed constant) is $T$. What is the initial volume according to the ideal gas law?

I'm not sure what you are asking me.

 

[kuruman]

I am asking you, if the pressure of the gas in the bubble matches the pressure at the bottom of the lake which is $p_{\text{bot.}}=p_{\text{atm.}}+\rho g h$ and the ideal gas law says $pV=nRT$, find an expression for the volume of the gas at the bottom. That's the volume that triples when the bubble rises to the surface.

 

[laser]

I am asking you, if the pressure of the gas in the bubble matches the pressure at the bottom of the lake which is $p_{\text{bot.}}=p_{\text{atm.}}+\rho g h$ and the ideal gas law says $pV=nRT$, find an expression for the volume of the gas at the bottom. That's the volume that triples when the bubble rises to the surface.

Is this what you mean?

 

[Steve4Physics]

View attachment 344858
I am a bit confused on the marking scheme as attached above.

P1V1 is a constant by Boyle's Law. If the volume increases by a factor of 3, then the pressure decreases by a factor of 3.

This means that the pressure at the top is 1/3 the pressure at the bottom, right? The pressure at the top is just atmospheric pressure (1E5 Pa), so the pressure at the bottom must be (3E5 Pa). 3E5 = Pressure due to water + Pressure due to air, so pressure due to water is just 2E5. So 2E5 = rho*g*h.

Is there a flaw in my understanding? Thanks!

No flaw. You are correct. The marking scheme incorrectly attributes the entire 3atm pressure to the water, which is wrong.

 

[kuruman]

View attachment 344861
Is this what you mean?

Yes. I see now that you had this result in post #1 when you wrote down
"So 2E5 = rho*g*h."
There is no flaw in your understanding.