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Heat transfer and mercury

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    The pressure on 0.01 litres of mercury is increased reversibly and isothermally
    from zero to 1000 atm at room temperature (293 K). Mercury has a coefficient
    of volume expansion β = 1.82 × 10−4 K−1 , and an isothermal compressibility
    κT = 4.02 × 10−11 Pa−1 . Note: 1 atm= 1.013 × 105 Pa. Assuming that the
    volume, V , changes very little, find
    (i) how much heat is transferred in the compression;
    (ii) the work done during the compression;
    (iii) the change in internal energy.

    2. Relevant equations
    [tex]T dS= C_p dT -\beta V dP [/tex]
    [tex] \oint \frac{\def\dbar{{\mathchar'26\mkern-12mu d}Q}
    \dbar}{T}=0[/tex]
    [tex]\def\dbar{{\mathchar'26\mkern-12mu d}Q}
    \dbar=C_v dT[/tex]

    3. The attempt at a solution
    I am attempting to find the final temp so I can implement: [tex]\def\dbar{{\mathchar'26\mkern-12mu d}Q}
    \dbar=C_v dT[/tex]

    Since the process is reversible, and under a complete cycle [tex] \oint \frac{\def\dbar{{\mathchar'26\mkern-12mu d}Q} \dbar}{T}=0[/tex] I set [tex]dS=0[/tex]. Getting [tex] C_p \int \frac{dT}{T}=\beta T V \int dP [/tex] Is this correct so far?
     
  2. jcsd
  3. Mar 18, 2009 #2

    Mapes

    User Avatar
    Science Advisor
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    Gold Member

    Your first and last equations doesn't look right. Check your units and your T's.
     
  4. Mar 18, 2009 #3
    I'd start with the first law.
     
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