 #1
mrjeffy321
Science Advisor
 875
 1
Technically, this isn’t really a home work problem, per se, but let’s try to solve it anyway, shall we.
We have a cylinder which at one end is closed off and at the other it is open. This cylinder has a total height H cm and diameter D cm (the real numbers aren’t all that important). Someone comes along and pushes the cylinder open end downward straight into a tub of water, trapping air inside as it descends into the murky deep. What is the pressure P_i inside the cylinder when the open ended bottom is X cm under the water and the outside pressure is P_o and water/air temperature of T?
So to help clarify the situation, I drew a picture (see attachment).
I think the first step to solving this to calculate the volume of the cylinder, so that is the area of the base times the height = H*pi*(D/2)^2 cubic cm
From this we can find out the number of moles of gas that is inside since the pressure inside starts out at the pressure above the water. Using the ideal gas law,
PV=nRT, we find that there (P_o * H*pi*(D/2)^2 )/RT moles of gas.
When the cylinder is pushed down into the water, the displaced water pushed back. So we can calculate the weight of water displaced would be its volume * its density (how lovely, water is 1 g/cc). So if we say the cylinder goes down X cm, then we calculate the volume of the displaced water the same as how we calculated the volume of the cylinder, so that is X*pi*(D/2)^2 cubic cm, which then gives us a back in the gas of (9.81)(X*pi*(D/2)^2 Newtons. This force is exerted over the entire area of the cylinder’s bottom, so that comes out to be ((9.81)(X*pi*(D/2)^2) / (pi*(D/2)^2), which simplifies to 9.81 * X N/cm^2 or 98100 * X Newtons per meter ^2 = 98100X Pascals = 98.1X kPa,
So now the total force is the new force of the water, plus the existing air pressure from above the water.
So now the gas has a pressure exerted on it, and using the ideal gas law again, the new volume should be,
V = nRT/P, and I’ll let you substitute it all in, since by now, everything is becoming highly symbolic and lengthy.
Since the volume of the gas gets smaller, the water rises up inside the tube to take its place, lessening ever, ever, ever so slightly the outside pressure, but this is so small we won’t worry about it.
So does this sound right?
In the end when the water is pushed up inside the tube, does that have an lessening effect on the pressure inside since that amount of water is no longer displaced by the gas, and this expands the air slightly, which intern displaces more water which increases the pressure, ……. Or no?
So I believe my end answer came out to be that when the cylinder is pushed X cm under the water, the pressure inside the tube is the pressure outside, P_o + 98.1X kPa,
Sound right?
I actually didn’t intend to solve this all here myself, but I was on a roll. Now I just need it confirmed.
We have a cylinder which at one end is closed off and at the other it is open. This cylinder has a total height H cm and diameter D cm (the real numbers aren’t all that important). Someone comes along and pushes the cylinder open end downward straight into a tub of water, trapping air inside as it descends into the murky deep. What is the pressure P_i inside the cylinder when the open ended bottom is X cm under the water and the outside pressure is P_o and water/air temperature of T?
So to help clarify the situation, I drew a picture (see attachment).
I think the first step to solving this to calculate the volume of the cylinder, so that is the area of the base times the height = H*pi*(D/2)^2 cubic cm
From this we can find out the number of moles of gas that is inside since the pressure inside starts out at the pressure above the water. Using the ideal gas law,
PV=nRT, we find that there (P_o * H*pi*(D/2)^2 )/RT moles of gas.
When the cylinder is pushed down into the water, the displaced water pushed back. So we can calculate the weight of water displaced would be its volume * its density (how lovely, water is 1 g/cc). So if we say the cylinder goes down X cm, then we calculate the volume of the displaced water the same as how we calculated the volume of the cylinder, so that is X*pi*(D/2)^2 cubic cm, which then gives us a back in the gas of (9.81)(X*pi*(D/2)^2 Newtons. This force is exerted over the entire area of the cylinder’s bottom, so that comes out to be ((9.81)(X*pi*(D/2)^2) / (pi*(D/2)^2), which simplifies to 9.81 * X N/cm^2 or 98100 * X Newtons per meter ^2 = 98100X Pascals = 98.1X kPa,
So now the total force is the new force of the water, plus the existing air pressure from above the water.
So now the gas has a pressure exerted on it, and using the ideal gas law again, the new volume should be,
V = nRT/P, and I’ll let you substitute it all in, since by now, everything is becoming highly symbolic and lengthy.
Since the volume of the gas gets smaller, the water rises up inside the tube to take its place, lessening ever, ever, ever so slightly the outside pressure, but this is so small we won’t worry about it.
So does this sound right?
In the end when the water is pushed up inside the tube, does that have an lessening effect on the pressure inside since that amount of water is no longer displaced by the gas, and this expands the air slightly, which intern displaces more water which increases the pressure, ……. Or no?
So I believe my end answer came out to be that when the cylinder is pushed X cm under the water, the pressure inside the tube is the pressure outside, P_o + 98.1X kPa,
Sound right?
I actually didn’t intend to solve this all here myself, but I was on a roll. Now I just need it confirmed.
Attachments

7 KB Views: 628
Last edited: