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Basic Thermodynamics; Change in U at Constant Pressure

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the ΔU° at 25°C for the following reaction at constant pressure:

    C[itex]_{2}[/itex]H[itex]_{2}[/itex] (g) + 5/2O[itex]_{2}[/itex] (g) [itex]\rightarrow[/itex] 2CO[itex]_{2}[/itex] (g) + H[itex]_{2}[/itex]O (g) ΔH° = -1299.5kJ

    2. Relevant equations

    ΔU = Q ± W
    PV = nRT
    W = PΔV

    3. The attempt at a solution

    Since we have constant pressure, ΔH° = Q, and since we are at standard conditions P = 1atm = 101.3kPa. Additionally T = 298K (which is given anyway). I try and find the work done on the system by finding the reduction in volume;

    ΔV = ΔnRT/P

    ΔV = (0.5 x 8.31 x 298)/101.3 = 12.2L

    W[itex]_{on system}[/itex] = 101.3kPa * 12.2L = 1238.2kJ

    ΔU = -1299.5kJ + 1238.2kJ = -61.3kJ

    However, the answer is -1298.3kJ.

    Interestingly, this answer is what I would get if I had accidently calculated the W[itex]_{on system}[/itex] was in joules and forgotten to convert - but I don't see how this can be the case - I'm dealing with kPa not Pa as my unit for pressure. Any help would be much appreciated!
     
  2. jcsd
  3. Jun 1, 2012 #2
    That is because ##E_{\mbox{int}}=Q+W## applies when work is done on the system.
     
  4. Jun 1, 2012 #3
    Work is being done on the system in this case; the pressure and temperature is constant, and so a decrease in the amount of gas results in a decrease of volume of the system. That is, the surrounding constant pressure is doing work in compressing the system.

    Either way, I have found my error - it was simply an error in dimensional analysis! 1kPa * 1L = 1J, not kJ! Whoops!
     
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