Calculate the water level in a pipe.

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To calculate the water level in a closed iron pipe submerged in water, the pressure equilibrium between the water and the air inside the pipe must be established. Using Boyle's Law, the relationship between pressure and volume is applied, where the initial pressure (p1) is the barometric pressure and the final pressure (p2) is determined by hydrostatic pressure. The volume of air inside the pipe is calculated using the formula v1 = πr²h1, and the height of the water level (h) can be derived from the relationship between the two volumes. The final calculation yields a water level of approximately 5.2 cm. This method effectively combines principles from both physics and chemistry to solve the problem.
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This was a question on my last exam for physical chemistry. There were no examples in the book as to how to do this question and I think it pertains to physics. I would like to know how to do this in the event that it appears on the final.

"An iron pipe 2 m long and closed at one end is lowered vertically into water until the closed end is flush with the water surface. Calculate the height h of the water level in the pipe. Additional data: 25 degrees Celsius, diameter of pipe = 3 in, density of water is 1 x 10^3 kgm^-3, barometric pressure is 100,000 Pa = 10 hydrostatic head of water. Neglect the effect of water vapor pressure."

Thanks
 
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The level of the water inside the pipe will be such that the water pressure and air pressure at the interface are equal to each other.
 
Boyle's Law applies here: p1v1=p2v2

I am assuming barometric pressure is p1
p2 is calculated by hydrostatic pressure
p2=p1+density*g*height

v1 = pi*r2*h1 (convert units)
v2= p1v1/p2

h2=v1/(pi*r2)

I get 5.2 cm
 
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