Final answer:
The question asks for the rms velocity of a rotating rigid diatomic molecule at a temperature of 300 K with a given moment of inertia. The calculation uses the equipartition theorem and rotational kinetic energy equations to establish the relationship, however, additional information is needed to complete the calculation.
Step-by-step explanation:
The question concerns the calculation of the root-mean-square (rms velocity) of the rotation of a rigid diatomic molecule when its temperature and moment of inertia are known. To calculate the angular velocity (ωrms), we apply the equipartition theorem, which states that each degree of freedom of a diatomic molecule contributes ½kT of energy, where k is the Boltzmann constant and T is the temperature. For diatomic gases at room temperature, which have two rotational degrees of freedom, the total rotational kinetic energy (2 x ½kT) is equally distributed over these two degrees. Additionally, diatomic molecules at this temperature do not yet exhibit vibrational motion, so it is not included in this energy consideration.
The formula used is the rotational kinetic energy equation, ER = ½Iωrms^2, set equal to the energy per degree of freedom, where I is moment of inertia. Since we have two degrees of freedom, the equation can be written as:
Iωrms^2 = kT
When we solve for ωrms, we get:
ωrms = √kT/I
Using the given values, T = 300 K and I = 2.1 x 10−39 g·cm², and knowing that k = 1.38 x 10−23 J/K, we perform the calculation to find the correct angular velocity.
However, to provide a complete answer, one would require additional information or context that is not given in the question, such as the explicit rms speed value or the mass of the diatomic molecule, to complete the calculation.