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

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(D) For an ideal gas,the speeds are defined as:${v_{rms}} = \sqrt{\frac{3kT}{m}}$,${v_p} = \sqrt{\frac{2kT}{m}}$,and $\bar{v} = \sqrt{\frac{8kT}{\pi m

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Both $(b)$ and $(c)$

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(D) For an ideal gas,the speeds are defined as:${v_{rms}} = \sqrt{\frac{3kT}{m}}$,${v_p} = \sqrt{\frac{2kT}{m}}$,and $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$.Comparing these values:${v_p} = \sqrt{2} \sqrt{\frac{kT}{m}} \approx 1.414 \sqrt{\frac{kT}{m}}$$\bar{v} = \sqrt{\frac{8}{3.14}} \sqrt{\frac{kT}{m}} \approx 1.596 \sqrt{\frac{kT}{m}}$${v_{rms}} = \sqrt{3} \sqrt{\frac{kT}{m}} \approx 1.732 \sqrt{\frac{kT}{m}}$Thus,${v_p} For the average kinetic energy:${E_{av}} = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \frac{3kT}{m} \right) = \frac{3}{2} kT$.Since ${v_p}^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.Substituting this into the energy equation:${E_{av}} = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$,which confirms option $(b)$ is correct.Therefore,both $(b)$ and $(c)$ are correct.

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 a molecule is $m$. Then

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