Calculation of enthelpy and internal energy or air in chemical equilibrium

[roldy]

Homework Statement

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.

$$K_{p,O_2}=12.19 atm$$
$$K_{p,N_2}=0.7899*10^{-4}atm$$

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

$$(\Theta_v)_{N_2}=3390 K$$
$$(\Theta_v)_{O_2}=2270 K$$

Homework Equations

$$(1)h=\Sigma ^n_{i=1}c_ih_i + \Sigma ^n_{i=1}c_i \Delta h_{f_i}$$

$$c_i=X_i\frac{\mu_i}{\mu}$$
$$X_i=\frac{P_i}{P}$$

$$R_{air}=\frac{R_u}{\mu_{air}}$$
$$\mu_{air}=X_{O_2}*\mu_{O_2} + X_O*\mu_O + X_N*\mu_N + X_{N_2}*\mu_{N_2}$$

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

Monatomic Gas:
$$h=5/2RT$$
$$=e_{sens}=3/2RT$$

The Attempt at a Solution

The weight of each species is as follows:
$$O=16kg/(kg*mol)$$
$$N=14kg(kg*mol)$$
$$O_2=32kg/(kg*mol)$$
$$N_2=28kg/(kg*mol)$$

I then found what the gas constant was for each species with Ru=8314 J/(kg*K).
$$R_O=519.625$$
$$R_N=593.8571$$
$$R_{O_2}=259.8125$$
$$R_{N_2}=296.9286$$

I then calculated the sensible internal enthalpy for each species.
$$e_{sense,O}=3/2(519.625)(4500)=3.5075*10^6$$
$$e_{sense,N}=3/2(593.8571)(4500)=4.0085*10^6$$
$$e_{sense,O_2}=3/2(259.8125)(4500) + (259.8125)(4500) +\frac{\frac{2270}{4500}}{e^{\frac{2270}{4500}}-1}(259.8125)(4500)=3.8218*10^6$$
$$e_{sense,N_2}=3/2(296.9286)(4500) + (296.9286)(4500) +\frac{\frac{3390}{4500}}{e^{\frac{3390}{4500}}-1}(296.9286)(4500)=4.2359*10^6$$

Next the enthalpy per unit mass of each species was calculated.
$$h_O=5/2(519.625)(4500)=5.8458*10^6$$
$$h_N=5/2(593.8571)(4500)=6.6809*10^6$$
$$h_{O_2}=3.8218*10^6+259.8125(4500)=4.9910*10^6$$
$$h_{N_2}=4.2359*10^6+296.9286(4500)=5.5721*10^6$$

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 $$c_i$$. I'm not sure how the equilibrium constants work into this problem.

 

[nate808]

I also do not see where the given values for the enthalpies of formation and vibrational temperatures come into play. Any help would be appreciated.

Thank you for your question. To calculate the enthalpy and internal energy per unit mass of the mixture, we need to first determine the composition of the mixture. This can be done using the equilibrium constants provided for O2 and N2.

Using the ideal gas law, we can calculate the partial pressures of O2 and N2:
P_{O_2} = K_{p,O_2} * P = 12.19 * 0.1 = 1.219 atm
P_{N_2} = K_{p,N_2} * P = 0.7899 * 10^{-4} * 0.1 = 7.899 * 10^{-6} atm

Next, we can use the given values for the enthalpies of formation to calculate the enthalpy of each species at 4500 K:
h_{O_2} = (3/2)RT + (\Delta H_f^o)_O = (3/2)(259.8125)(4500) + 2.47 * 10^8 = 3.8219 * 10^6 J/kg
h_{N_2} = (3/2)RT + (\Delta H_f^o)_N = (3/2)(296.9286)(4500) + 4.714 * 10^8 = 4.2365 * 10^6 J/kg

To calculate the enthalpy and internal energy of the mixture, we first need to determine the mole fractions of each species:
X_{O_2} = P_{O_2}/P = 1.219/0.1 = 0.01219
X_{N_2} = P_{N_2}/P = 7.899 * 10^{-6}/0.1 = 7.899 * 10^{-5}

Next, we can use the mole fractions to calculate the molecular weight and specific gas constant of the mixture:
\mu_{air} = X_{O_2} * \mu_{O_2} + X_{N_2} * \mu_{N_2} = (0.01219)(32) + (7.899 * 10^{-