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Pressure difference between two bulbs measured by a mercury manometer

  1. Aug 12, 2012 #1
    The setup is a particular gas thermometer constructed of two gas containing bulbs, each of which is put into a water bath on adjacent sides. The pressure difference between the two bulbs is measured by a mercury manometer. Appropriate reservoirs maintain constant gas volume in the two bulbs. There is no difference in pressure when both baths are at the triple point of water. The pressure difference is 120mmHG when one bath is at the triple point and the other is at the boiling point of water. Finally, the pressure difference is 90.0 mm Hg when one bath is at the triple point and the other is at an unknown temperature to be measured. Find the unknown temperature.

    The solution says "For either container we can write piVi = niRTi. We are told that ni and Vi are constants. Then Δp = AT1 - BT2, where A and B are constants." I need help formulating the constants A and B, and an explanation. Thanks!!!!
     
    Last edited: Aug 12, 2012
  2. jcsd
  3. Aug 12, 2012 #2
    [itex]p_iV_i = n_iRT_i[/itex] implies [itex]p_i = \frac {n_iR} {V_i} T_i[/itex], [itex]\frac {n_iR} {V_i} = c_i = const[/itex]
     
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