Physically, why is a gas molecule's speed $v$ more likely to be near $v_{p}$ than near $2 v_{p}$ ?
A) There are fewer quantum states for larger speeds than there are at speeds near $v_{p}$.
B) The Boltzmann factor is larger at $\mathbf{v}_{\mathbf{p}}$ than at $\mathbf{2} \mathbf{v}_{\mathbf{p}}$.
C) We know that $\frac{1}{2} m\left[v^{2}\right]_{\text {avg }}=\frac{3}{2} k_{B} T$, and $\mathrm{v}_{\mathrm{p}}$ is closer to $\left[v_{\text {avg }}^{2}\right]^{1 / 2}$ than $2 \mathrm{v}_{\mathrm{p}}$ is.

Question

Quizplus · Accepted Answer

The Boltzmann factor, which describes the distribution of kinetic energies (and thus speeds) among particles in a gas, decreases exponentially with increasing energy. Since the kinetic energy is proportional to the square of the speed ($v^2$), a speed of $2v_p$ corresponds to four times the kinetic energy of $v_p$, making the Boltzmann factor significantly smaller at $2v_p$ than at $v_p$, and thus, speeds near $v_p$ are statistically more likely.

Physically, Why Is a Gas Molecule's Speed vvv More Likely to Be Near

Physically, Why Is a Gas Molecule's Speed $v$ More Likely to Be Near