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Ideal Gasses, Double-Layered floating objects, and immersion in oil

  1. Mar 27, 2006 #1

    I'm absolutely confused on this number 3. I've attempted this question but there must be something I'm missing out which makes a difference. An explanation would be nice. In order to find the displacement in part a wouldn't you need to know the pressure at 270 degrees? I used the ideal gas law and came down to Pf*Vf/Pi*Vi=Tf/Ti and I have everything except for the final pressure and final volume. Now I attempted to figure out the final pressure by adding 1 atm + 1.9738x atm figuring that the force/area of the spring plus the initial pressure would be the final pressure. Is it plus or is it minus? After that I tried the final volume as being 6 L + 0.01x. Again is that plus or minus? And then I plugged in those two terms into the equation and solved the quadratic to come out to the 43.15 cm which is between 10% and 100% off. I could use a little guidance on this one.

    3. An expandable cylinder has its top connected to a spring with force constant 2.00 103 N/m. The cylinder is filled with 6.00 L of gas with the spring relaxed at a pressure of 1.00 atm and a temperature of 20.0°C.

    (a) If the lid has a cross-sectional area of 0.0100 m2 and negligible mass, how high will the lid rise when the temperature is raised to T = 270°C? (In cm)

    I answered 43.14 cm and it says:
    Your answer differs from the correct answer by 10% to 100%.

    (b) What is the pressure of the gas at 270°C? (In Pa)


    As for this one I've completed part a yet I can't seem to figure out part b. I tried to say that since the Toil-Tair=40N I could have the Buoyant force be 40N. Then I'd have 40N=density of oil*g*V. Now I figured that the volume would be the same as the 1*10^-4 cubic meters that I figured out as part of part a since that is the displaced volume. I figured that the density of oil would then be 40,816 kg/m^3 but it was orders of magnitude off. As for part a where I was correct I found that the Buoyant force is 330N-250N=80N. Then I found that the Volume of water=80N/(g*density of water) which comes out to 8.16*10^-3m^3. Once I had that I knew that the density of the object would be mass over volume of water displaced since that is equal to the volume of the object. So it comes to 33.7 kg/0.0082m^3=4125 kg/m^3 which is correct. I could use help on part b.

    5. An object weighing 330 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 250 N. Immersed in oil, the object weighs 290 N.

    (a) Find the density of the object. kg/m3

    (b) Find the density of the oil. kg/m3

    I entered 40,816 kg/m^3 and it said:
    Your answer differs from the correct answer by orders of magnitude.


    Lastly I have 1 out of 3 on number 8 correct. Part a I am correct and the answer is 16.96 mm. I know that the ice cube has a mass of 0.0058 kg. I found that through multiplying the (density of ice*the volume of the ice cube)/(density of water*area of ice cube)=16.96 mm. Then for part b and c I am lost. I know that ethyl alcohol has a density of 0.806*10^3 kg/m^3. I know that I have to somehow apply archimedes principle to an object in two different layers. Any help would be appreciated.

    8. A cube of ice whose edge is 18.5 mm is floating in a glass of ice-cold water with one of its faces parallel to the water surface.

    (a) How far below the water surface is the bottom face of the block? 16.96 mm (correct)

    (b) Ice-cold ethyl alcohol is gently poured onto the water surface to form a layer 5.00 mm thick above the water. When the ice cube attains hydrostatic equilibrium again, what will be the distance from the top of the water to the bottom face of the block? (mm)

    (c) Additional cold ethyl alcohol is poured onto the water surface until the top surface of the alcohol coincides with the top surface of the ice cube (in hydrostatic equilibrium). How thick is the required layer of ethyl alcohol? (mm)
  2. jcsd
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