In the Maxwell-Boltzmann distribution of mole | vedclass

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(B) The most probable speed $(U_{M.P.})$ is given by the formula $U_{M.P.} = \sqrt{\frac{2RT}{M}}$.From this expression,it is clear that $U_{M.P.} \pr

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$U_{M.P.}$ at $T_2 > U_{M.P.}$ at $T_1$

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(B) The most probable speed $(U_{M.P.})$ is given by the formula $U_{M.P.} = \sqrt{\frac{2RT}{M}}$.From this expression,it is clear that $U_{M.P.} \propto \sqrt{T}$.As the temperature increases,the peak of the Maxwell-Boltzmann distribution curve shifts to the right,indicating a higher most probable speed.In the given graph,the peak for $T_2$ is at a higher velocity value compared to the peak for $T_1$,which implies $T_2 > T_1$.Therefore,$U_{M.P.}$ at $T_2 > U_{M.P.}$ at $T_1$.

In the Maxwell-Boltzmann distribution of molecular speeds shown below,which of the following is correct regarding the most probable speed $(U_{M.P.})$ at temperatures $T_1$ and $T_2$?

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