# Atmospheric pressure and water pressure - Boyle's law

• laser
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!

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?

MatinSAR
kuruman said:
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.

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.

MatinSAR
kuruman said:
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?

laser said:
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.

MatinSAR and laser
laser said:
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.

MatinSAR and laser

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