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Change in Enthalpy for Nitrogen at constant temp

  1. Jun 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate ΔH for 2 mols of Nitrogen for the following transformation:
    1 bar, 21°C → 200 bar, 21°C

    Given that molar Cp = 4R

    2. The attempt at a solution
    How am I supposed to find the ΔH? From what I learned, there can only be a change in enthalpy when there's a change in temperature.

    So the change in enthalpy will always be zero for an isothermal expansion.
  2. jcsd
  3. Jun 2, 2016 #2
    yes for an ideal gas enthalpy does not change if the temperature does not change.

    Perhaps its one of those trap-like tricky questions...
  4. Jun 2, 2016 #3

    James Pelezo

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    Gold Member

    I agree, that in order to have an Enthalpy of Rxn a temperature change needs to be measurable. However, if the process is adiabatic with an isothermal phase transition (that is, ΔH = 0) application of the system given to the 1st Law gives ΔU = ΔH - PΔV = (0) - PΔV = - PΔV = work done by the system on surroundings or surroundings on the system. For ΔV > 0 => Expansion Process => Exothermic Work or, ΔV < 0 => Compression Process => Endothermic Work. So while it is true ΔH in this case would be zero, the work done by the surroundings on the system would not be zero, and (in my humble opinion) 'work' would be the objective value in the problem.

    So, from the posted problem, I see an adiabatic compression of 2 moles of N2(g) at 1bar pressure (~1 atm) to 2 moles of N2(g) at 200 bar (~200 atm). Such a compression would most likely drive the N2(g-phase) => N2(l-phase) and would transition via adiabatic process (where enthalpy, ΔH = 0) and demonstrate a '0' temperature change. This condition warrants use of the expression defining 'Heat of Condensation/Vaporization for N2. That is ΔUg/l = ΔHg/l transition = mass of N2 x ΔHt, and ΔHt = published phase transition constant for N2. From Wiki (https://en.wikipedia.org/wiki/Nitrogen), ΔHv(N2) = 5.56 Kj/mole. So, for 2 moles molecular nitrogen = 2 Molar Volumes (2VmN2) undergoing phase transition (g →l) due to compression process => ΔU = ΔH - (- Work by Compression*) = ΔU = ΔH + (Work by Compression*) => ...

    * Work by Compression => Work done by the surroundings on the system => 'Endothermic' to the System. => Work = PΔV = P(Vfinal - Vinitial) => Work = P(0Vm - 2Vm) = - 2Vm. Substituting into ΔU = ΔH - (P(-2Vm) = ΔH + 2PΔV = ΔH + (Work done on system by compression) = (ΔH) + (m⋅ΔHv) = (0) + (2 moles N2)(5.56 Kj/mole) = + 11.12 Kj.
  5. Jun 2, 2016 #4
    So you see an in between adiabatic process and a phase change but the problem at least as stated in the OP implies just an isothermal compression.

    We got to know exactly how the problem is stated in his book to be sure.
  6. Jun 2, 2016 #5
    Enthalpy is a function solely of temperature only for an ideal gas. For a real gas in the non-ideal gas region (i.e., higher pressures), enthalpy is also a function of pressure. Certainly in the case of nitrogen, at pressures approaching 200 bars, the behavior of the gas will be non-ideal. So to get the change in enthalpy between the two states in this problem, the non-ideal gas behavior of the nitrogen needs to be taken into account.

    For a pure single-phase substance, the effect of pressure on molar enthalpy at constant temperature is given by the equation:
    $$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
    where V is the molar volume. Note that the right hand side of this equation is equal to zero for an ideal gas, but not for a real gas. To get the change in enthalpy for this problem, you need to integrate the right hand side of the equation with respect to P. This can't be done unless you have an equation of state for nitrogen (e.g., van der Waals). Do you have an equation of state with specific parameters for nitrogen?

    Thinking that the nitrogen can be liquified at 21 C and very high pressures does not make sense because the critical temperature of nitrogen is -147 C, and a gas cannot be liquified at temperatures above its critical temperature, no matter how much pressure is applied.
    Last edited: Jun 3, 2016
  7. Jun 4, 2016 #6
    Check out post #5 of this thread. It's an example of what @Chestermiller told you, and you will find it pretty useful. Of course, you should try it yourself with whatever equation of state you're using in class.
  8. Jun 4, 2016 #7

    James Pelezo

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    Gold Member

    Of course, always check the phase diagram, should of known better... duh! Thanks Chet, no excuses for this one. Anyways, looking back at the problem, it's clear that the system is experiencing extreme compressibility but not a phase change. Your note makes sense, it's a more rational approach. When I get more time I'll study it more, but am prep'n for summer classes starting Monday... 12 weeks to cover 24 chapters of Ebbing, what a rush, but it's fun... I like the summer classes... Good students...they work hard at making the grade... Will be back.. Later, JP
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