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Calculation of enthalpy and internal energy or air in chemical equilibrium

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider air in chemical equilibrium at 0.1 atm and T=4500 K. The chemical species are O2, O, N2, N (Ignore NO). Calculate the enthalpy and internal energy per unit mass of the mixture. Neglect electronic excitation in your calculations.

    [tex]K_{p,O_2}=12.19 atm[/tex]

    [tex](\Delta H_f^o)_O=2.47*10^8 J/(kg*mol)[/tex]
    [tex](\Delta H_f^o)_N=4.714*10^8 J/(kg*mol)[/tex]

    [tex](\Theta_v)_{N_2}=3390 K[/tex]
    [tex](\Theta_v)_{O_2}=2270 K[/tex]

    2. Relevant equations
    [tex](1)h=\Sigma ^n_{i=1}c_ih_i + \Sigma ^n_{i=1}c_i \Delta h_{f_i}[/tex]


    [tex]\mu_{air}=X_{O_2}*\mu_{O_2} + X_O*\mu_O + X_N*\mu_N + X_{N_2}*\mu_{N_2}[/tex]

    Diatomic Gas:
    [tex]e_{sens}=3/2RT + RT +\frac{\frac{\Theta_v}{T}}{e^{\frac{\Theta_v}{T}}-1}RT[/tex]

    Monatomic Gas:

    3. The attempt at a solution

    The weight of each species is as follows:

    I then found what the gas constant was for each species with Ru=8314 J/(kg*K).

    I then calculated the sensible internal enthalpy for each species.
    [tex]e_{sense,O_2}=3/2(259.8125)(4500) + (259.8125)(4500) +\frac{\frac{2270}{4500}}{e^{\frac{2270}{4500}}-1}(259.8125)(4500)=3.8218*10^6[/tex]
    [tex]e_{sense,N_2}=3/2(296.9286)(4500) + (296.9286)(4500) +\frac{\frac{3390}{4500}}{e^{\frac{3390}{4500}}-1}(296.9286)(4500)=4.2359*10^6[/tex]

    Next the enthalpy per unit mass of each species was calculated.

    This is where I get stuck. I believe I use equation (1) in some way but I do not see how since I am not given the partial pressures to find [tex]c_i[/tex]. I'm not sure how the equilibrium constants work into this problem.
  2. jcsd
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