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Calculation of work done

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Decomposition of 2 moles of NH4NO3 at 100 degrees C.calculate work done

    2. Relevant equations
    NH4NO3 (s)→N2O(g)+2H2O (g)
    work done= - RT delta n here R=universal gas constant T=temperature in kelvin and
    delta n=difference in number of moles of products and reactants.
    3. The attempt at a solution-I pluged in the values in formula work done= - RT delta n
    =- 8.314j per kelvin per mol×373 kelvin ×3 mol
    =- 9303.366 j
    I took delta n =3 because i think delta number of moles in solid phase is taken as zero.Even if i don't number of moles in solid phase =0 rather I Take number of moles in solid phase = 1, I will get delta n =2 and then

    work done= - RT delta n
    =- 8.314 j per kelvin per mol ×373 kelvin ×2 mol
    =- 6202.244 j
    But according to my textbook answer should be - 18.61kj
    please help.
     
  2. jcsd
  3. Dec 1, 2014 #2

    Borek

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    Staff: Mentor

    That's per decomposition of one mole of the nitrate. Now reread the question.

    We ignore the initial volume of the solid as typically STP it is orders of magnitude smaller than the volume of gases produced (in the case of NH4NO3 it will be around 47 mL vs 66 L). It is often a good approximation, but not always.

    Question is poorly written. 100°C doesn't guarantee water is in the gaseous form, for that pressure must be lower than 1 atm.
     
  4. Dec 1, 2014 #3
    can you please tell me why number of moles in solid phase is taken as zero?[/QUOTE]
     
  5. Dec 1, 2014 #4

    Borek

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    Staff: Mentor

    I already did - that's because we can ignore its volume.

    Work done is not RTΔn, but PΔV. From the ideal gas equation we know PV=nRT, so the volume of the gas produced is V=ΔnRT/P. In general ΔV is not just the volume of the gas produced, actually it is Vfinal-Vinitial. In the decomposition of ammonium nitrate, initial volume is that of a solid - around 47 mL per two moles (assuming density of the solid to be 1.7 g/mL). Final volume is orders of magnitude higher (measured in tens of liters), so we just ignore initial volume of the solid, as it is thousand times lower

    In other words, as Vfinal >> Vinitial, Vfinal-Vinitial ≈ Vfinal.
     
  6. Dec 1, 2014 #5
    In my textbook Work done is given as RTΔn.I think it is like this pv=nRT and work done=p delta v so at constant temperature and pressure p delta v=delta n RT so work done=delta n RT.I want to ask can we take change in volume of solid=final volume in every case?
     
    Last edited: Dec 1, 2014
  7. Dec 1, 2014 #6

    Borek

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    Staff: Mentor

    Yes, that's what I wrote.

    No idea what you mean. Volume of solid is not a final volume, I never stated anything like that. Please reread my explanation, I think I addressed what is going on quite clearly.
     
  8. Dec 1, 2014 #7
    If we have reactant which is solid ,can we just ignore initial volume of the solid in every case, is it thousand times lower than final volume in everycase? so that Vfinal >> Vinitial, Vfinal-Vinitial ≈ Vfinal.l.
     
  9. Dec 1, 2014 #8

    Borek

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    Staff: Mentor

    Depends on the pressure - the higher the pressure, the lower the gas volume. If the pressure is high enough gas volume can be so low condition V
    final >> Vinitial is no longer meet.
     
  10. Dec 1, 2014 #9
    so can you please tell me range,i mean how much high the pressure should be so that Vfinal >> Vinitial is no longer meet?It will help me to solve problems.
     
  11. Dec 1, 2014 #10

    Borek

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    Staff: Mentor

    In general, unless the conditions are somewhat exotic (very high pressures - hundreds of atm, very low temperatures - like -100°C), you don't have to worry about the initial solid volume.
     
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