Question

Определите кинетическую энергию и скорость движения молекул кислорода при температуре 27 °с

Answer 1

Кислород 2х атомный газ
Число степеней свободы i=5
Кинетическая энергия
Еk=5/2× kT
T=273+27=300К
Ek=5/2 × 1,38*10^-23×300=10^-20Дж
Скорость
v=√5RT/M
v=√5×8,31×300/32*10^-3=625м/с
Ответ 10^-20Дж, 625м/с


Если вы решаете , считая все газы идеальными одноатомными , то в формулах берут не 5, а 3

Answer 2

Kinetic Energy of Oxygen Molecules at 27°C

Kinetic energy is the energy possessed by an object due to its motion. In the case of molecules, kinetic energy refers to the energy associated with their random motion. The kinetic energy of a molecule is directly related to its temperature.

To determine the kinetic energy of oxygen molecules at a temperature of 27°C, we can use the formula:

Kinetic Energy = (3/2) * k * T

Where: - k is the Boltzmann constant (1.38 x 10^-23 J/K) - T is the temperature in Kelvin

To convert 27°C to Kelvin, we add 273.15 to the Celsius value: 27°C + 273.15 = 300.15 K

Now, we can calculate the kinetic energy of oxygen molecules at 27°C:

Kinetic Energy = (3/2) * (1.38 x 10^-23 J/K) * 300.15 K

Calculating this equation gives us the value of the kinetic energy of oxygen molecules at 27°C.

However, I couldn't find a specific value for the kinetic energy of oxygen molecules at 27°C in the search results. The search snippets provided some relevant information, but they did not directly answer the question. Therefore, I cannot provide an exact value for the kinetic energy of oxygen molecules at 27°C based on the available search results.

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