Question

# A nondegenerate ideal gas of diatomic molecules with a kilomolar mass of 2 kg/kmol and a characteristic rotational temperature of 86 K is adsorbed on the walls of a container, where the binding energy is 0.02 eV. The adsorbed molecules move freely on the walls, and their rotation is confined to the plane of the walls. Calculate the surface density of adsorbed molecules at 12 K if the gas pressure is 103 Pa! What result would you get at 68 K and the same pressure?

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## Answer to a math question A nondegenerate ideal gas of diatomic molecules with a kilomolar mass of 2 kg/kmol and a characteristic rotational temperature of 86 K is adsorbed on the walls of a container, where the binding energy is 0.02 eV. The adsorbed molecules move freely on the walls, and their rotation is confined to the plane of the walls. Calculate the surface density of adsorbed molecules at 12 K if the gas pressure is 103 Pa! What result would you get at 68 K and the same pressure?

To calculate the surface density of adsorbed molecules, we can use the Langmuir adsorption isotherm equation: Γ = $P * A$ / $k * T$ where: Γ is the surface density of adsorbed molecules in mol/m², P is the gas pressure in Pa, A is the binding energy per molecule in J, k is the Boltzmann constant $1.38 x 10^-23 J/K$, T is the temperature in Kelvin. First, let's convert the binding energy from eV to J: 1 eV = 1.602 x 10^-19 J So, the binding energy is 0.02 eV * 1.602 x 10^-19 J/eV = 3.204 x 10^-21 J. Now, let's calculate the surface density at 12 K: Γ₁ = $103 Pa * 3.204 x 10^-21 J$ / $1.38 x 10^-23 J/K * 12 K$ Γ₁ ≈ 1.87 x 10^5 mol/m² Next, let's calculate the surface density at 68 K: Γ₂ = $103 Pa * 3.204 x 10^-21 J$ / $1.38 x 10^-23 J/K * 68 K$ Γ₂ ≈ 5.26 x 10^4 mol/m² Therefore, at 12 K and a gas pressure of 103 Pa, the surface density of adsorbed molecules is approximately 1.87 x 10^5 mol/m², while at 68 K and the same pressure, the surface density is approximately 5.26 x 10^4 mol/m².
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