Let bar{v},v_{rms} and v_p respectively denot | vedclass

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(D) The expressions for the speeds are:$v_{rms} = \sqrt{\frac{3kT}{m}}$$\bar{v} = \sqrt{\frac{8kT}{\pi m}} \approx \sqrt{\frac{2.55kT}{m}}$$v_p = \sqr

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the average kinetic energy of the molecules is $\frac{3}{4}(mv_p^2)$

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(D) The expressions for the speeds are:$v_{rms} = \sqrt{\frac{3kT}{m}}$$\bar{v} = \sqrt{\frac{8kT}{\pi m}} \approx \sqrt{\frac{2.55kT}{m}}$$v_p = \sqrt{\frac{2kT}{m}}$Comparing these,we get $v_p For the average kinetic energy of a molecule:$E_k = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \sqrt{\frac{3kT}{m}} \right)^2 = \frac{3}{2} kT$.Since $v_p^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.Substituting this into the expression for $E_k$:$E_k = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$.

Let $\bar{v}$,$v_{rms}$ and $v_p$ respectively denote the mean speed,root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of the molecule is $m$. Then:

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