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Thermodynamics of mercury thermometers

  1. Aug 27, 2015 #1
    • Please don't ignore the template.
    The space above the mercury column in a thermometer ordinarily is evacuated, but due to faulty manufacture, a particular thermometer has a pressure of 2 mmHg of air in this space when the whole thermometer is immersed in a bath at 0 degrees Celsius. Calculate the pressure of the air when the whole thermometer is immersed in a bath at 10 degrees Celsius. At 0 degrees Celsius the length of air space is 10 cm and at 10 degrees Celsius the length of air space is 2 cm.

    I started out by using p = rho * g * h. For 0 degrees Celsius I got that rho = 212.31 Pa. I'm not sure that rho would be the same for 100 degrees Celsius though, because the pressure and length are changing. If rho is different, how would I go about solving for it to fill into the formula for 100 degrees celsius?
     
  2. jcsd
  3. Aug 27, 2015 #2

    SteamKing

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    In the formula P = ρ ⋅ g ⋅ h, ρ is the mass density of the fluid, so it does not have units of pascals. :frown:

    Since the manufacturing of the thermometer was faulty and allowed air to enter, it seems the problem is asking you to consider the effect that temperature has on this trapped air, and how this affects the accuracy of the device. :wink:
     
  4. Aug 28, 2015 #3
    So the pressure of the air equals the pressure of the mercury. If the change in length is 10cm - 2 cm = 8cm, then the mercury would have increased 80 mmHg. Add this to the original 2 mmHg and you get 82 mmHg at 10 degrees Celsius. Am I thinking about this correctly?
     
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