Consider an equimolar mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\) gases at \(800 \mathrm{~K}\) and a total pressure of \(0.5\) atm. For a path length of \(1.2 \mathrm{~m}\), determine the emissivity of the gas.

Short Answer

Expert verified
Question: Calculate the molar concentration of an equimolar mixture of CO2 and O2 gases, given the following conditions: total pressure of 0.5 atm, temperature of 800 K, and a path length proportional to a volume of 1.2 m3. Answer: The molar concentration for each gas (CO2 and O2) is 4.573 mol.

Step by step solution

01

Calculate the molar concentration of each gas

Since the mixture is equimolar, the molar concentration of each gas is equal. Let's first calculate the total moles of gas within the given total pressure. We can calculate the moles of gas using the ideal gas law: \(PV = nRT\) Where: \(P = 0.5 \ \text{atm}\) \(V = 1.2 \ \text{m}^3\) (Assuming the path length is proportional to volume) \(T = 800 \ \text{K}\) \(R = 0.0821 \ \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\) First, we need to convert the volume to liters: \(V (L) = 1.2 \ \text{m}^3 \times \frac{1000 \ \text{L}}{1 \ \text{m}^3} = 1200 \ \text{L}\) Now, we can solve for \(n\) (the total moles): \(n = \frac{PV}{RT} = \frac{(0.5 \ \text{atm})(1200 \ \text{L})}{(0.0821 \ \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}})(800 \ \text{K})} = 9.145 \ \text{mol}\) The molar concentration for each gas (n') is half of the total moles: \(n' = \frac{9.145 \ \text{mol}}{2} = 4.573 \ \text{mol}\)
02

Calculate partial pressures of each gas

Now we need to find the partial pressure of each gas, using the mole fraction: For an equimolar mixture, the mole fraction is equal to \(\frac{1}{2}\) for each gas. Partial pressure\( \ (P_{i}) = \text{mole fraction}_i \times P_{\text{total}}\) For both \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\) gases: \(P_{\mathrm{CO}_{2}} = P_{\mathrm{O}_{2}} = \frac{1}{2} \times 0.5 \ \text{atm} = 0.25\) atm
03

Determine the weighted sum of emissivities

Now, we need to determine the individual emissivities for each gas \((\varepsilon_i)\) and combine them into a single emissivity value for the mixture using the weighted sum of the emissivities: \(\varepsilon_\text{total} = \frac{\sum_{i=1}^N \varepsilon_i P_i}{\sum_{i=1}^N P_i}\) Where \(P_i\) is the partial pressure of each gas and \(N\) is the number of gases in the mixture. Unfortunately, we cannot compute the individual emissivities because the values are not provided, and these values are based on experimental measurements. Typically, one would have to consult a table or database to find the corresponding emissivity values for each gas at the given conditions (temperature, pressure, and path length). If we had the emissivities for each gas, we would substitute them in the above equation and calculate the result. However, this is beyond the scope of the current exercise, as determining these values requires specific experimental data.

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Most popular questions from this chapter

What does the view factor represent? When is the view factor from a surface to itself not zero?

Consider two infinitely long concentric cylinders with diameters 20 and \(25 \mathrm{~cm}\). The inner surface is maintained at \(700 \mathrm{~K}\) and has an emissivity of \(0.40\), while the outer surface is black. If the rate of radiation heat transfer from the inner surface to the outer surface is \(2400 \mathrm{~W}\) per unit area of the inner surface, the temperature of the outer surface is (a) \(605 \mathrm{~K}\) (b) \(538 \mathrm{~K}\) (c) \(517 \mathrm{~K}\) (d) \(451 \mathrm{~K}\) (e) \(415 \mathrm{~K}\)

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