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Thermodynamics: dropping iron into thermos of nitrogen at boiling point

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    You come into lab one day and find a well insulated 2000mL thermos bottle containing 500mL of boiling liquid nitrogen. The remainder of the thermos has nitrogen gas at a pressure of 1.0atm. The gas and liquid are in thermal equilibrium. While waiting for lab to start, you notice a piece of iron on the table with "197g" written on it. Just for fun, you drop the iron into the thermos and seal the cap tightly so that no can can escape. After a few seconds have passed, what is the pressure inside the thermos? The density of liquid nitrogen is 810kg/m^3


    2. Relevant equations
    PV=nRT
    Enthalpy of Vaporization (Nitrogen) = 5.577kJ/mol
    specific heat (iron) = 0.45J/gK
    Q=mass*specific heat*Δtemperature
    Room Temperature = 21C = 293K
    Boiling Point Nitrogen = 77.36K

    3. The attempt at a solution

    1.) Find energy value of the iron

    using iron's specific heat value, I can find out how much Q iron brings to the system.

    Knowing that the final temperature of the iron is going to be 77.36K because of how much energy it takes to vaporize all of the liquid nitrogen, i set up iron's Q equation:

    Q(iron) = 197g*0.45J/gK*(77.36K-293K) = -19116.48J

    this pretty much means that the iron is bringing in 19116.48J into the system.

    2.) 19116.48J is done on liquid nitrogen

    • temperature will not change
    • will try to change phase into vapor
    • know that enthalpy of vaporization for nitrogen is 5.577kJ/mol

    this 5.577kJ/mol value, to me, seems to say that nitrogen changes from a liquid to a gas at the rate of 1 mol for every 5.577kJ done on it.

    using this logic, I calculate how many mols of the liquid nitrogen turns into gas.

    5577J/mol = 19116.48J/x mol

    using a calculator, x = 3.42

    3.) 3.42 mols of nitrogen gas enters the system

    now that we are talking gasses, I may use the ideal gas law PV=nRT

    1atm = 101325pa

    for the initial state of the system
    P = 101325pa
    V = 0.0015m^3
    T = 77.36K
    n = (after calculating) = 0.236 mols

    for the final state of the system
    V=
    since 3.42 mols of liquid nitrogen also leaves the system
    3.42*(atomic mass of nitrogen is 14g/mol) = 47.88g
    density of liquid nitrogen is 810kg/m^3

    810000g/m^3 = 47.88g/(x)m^3

    x = 5.91X10^(-5)

    so V = (0.0015+0.0000591)m^3 = 0.0015591
    n = (0.236+3.42)mol = 3.656mol
    T = 77.36K
    so P is 1508284.42pa or ≈ 15atm which is the book's answer.

    Does it look like I solved this correctly? I never got the answer right until now!
     
  2. jcsd
  3. May 14, 2012 #2
    Looks good to me. You didn't really need to find the additional gas volume that was made available by the nitrogen that boiled. It's just such a tiny volume.
     
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