The relation between the average kinetic energy of a molecule, 1/2 mv^2= 3/2kT.
m is the massof the molecule; v is the average velocity; k is 1.38x10^-23J/K; T is the absolute temperature. 1J=1 kg m^2s^-2. Calculate the average velocity of a nitrogen dioxide molecule in the atmosphere at 27.0 degrees Celsius
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2017-08-30 4:12 pm
回答 (2)
2017-08-30 4:48 pm
Molar mass of NO₂, M = (14.01 + 16.00×2) g/mol = 46.01 g/mol = 0.04601 kg/mol
Avogadro constant, L = 6.022 × 10²³ /mol
Mass of an NO₂ molecule, m = M / L = (0.04601 kg/mol) / (6.022 × 10²³) = 7.64 × 10⁻²⁶ kg
(1/2) m v² = (3/2) k T
v² = 3 k T/ m
v = √[3 k T/ m]
Average velocity, v = √[3 × (1.38 × 10⁻²³ kg m² s⁻² K⁻¹) × (273.2 + 27.0 K) / (7.64 × 10⁻²⁶ kg)] = 403 m s⁻¹
Avogadro constant, L = 6.022 × 10²³ /mol
Mass of an NO₂ molecule, m = M / L = (0.04601 kg/mol) / (6.022 × 10²³) = 7.64 × 10⁻²⁶ kg
(1/2) m v² = (3/2) k T
v² = 3 k T/ m
v = √[3 k T/ m]
Average velocity, v = √[3 × (1.38 × 10⁻²³ kg m² s⁻² K⁻¹) × (273.2 + 27.0 K) / (7.64 × 10⁻²⁶ kg)] = 403 m s⁻¹
2017-08-30 4:43 pm
1/2 m <v^2> = 3/2 kT
<v^2> = 3 k T / m
= 3 * 1.38*10^-23 kg m^2/s^2 /K * 300K / (4.65*10^-26 kg)
= 2.67*10^6 m^2/s^2
v_rms = sqrt(2.67*10^6 m^2/s^2) = 517 m/s
<v^2> = 3 k T / m
= 3 * 1.38*10^-23 kg m^2/s^2 /K * 300K / (4.65*10^-26 kg)
= 2.67*10^6 m^2/s^2
v_rms = sqrt(2.67*10^6 m^2/s^2) = 517 m/s
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