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MV². 2RT dN M F(V)dV =5 N = 47(- 3:2y² exp(- `2TRT AP(= Find the Vm value by considering the following curve. 20 x 10 IS x 10 10 x 10 1500 K 5x 10 Vm Vm 10002 (0002

MV². 2RT dN M F(V)dV =5 N = 47(- 3:2y² exp(- `2TRT AP(= Find the Vm value by considering the following curve. 20 x 10 IS x 10 10 x 10 1500 K 5x 10 Vm Vm 10002 (0002

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M MV² dN = 47(- N )3/2y² exp(- 2 RT F(V)dV = -)dv -)dV 2RT Find the Vm value by considering the following curve. ax 10 20 x 10 298 K 15 x 10 10 x 10 1500 K 5x 10 Vm Vm | 10002
Transcribed Image Text:M MV² dN = 47(- N )3/2y² exp(- 2 RT F(V)dV = -)dv -)dV 2RT Find the Vm value by considering the following curve. ax 10 20 x 10 298 K 15 x 10 10 x 10 1500 K 5x 10 Vm Vm | 10002
Expert Solution
Step 1

Here Vm is the most probable speed as shown in the graph

most probable velocity physically means that the maximum number of particles will possess this velocity, as can seen from the graph.

from the distribution function provided

F(V)dV=dNN=4πM2πRT3/2 V2 exp-MV22RTdVFrom the concept of maxima and minima, at V=Vm the derivative of the distribution function will be 0substituting M2πRT=A and M2RT=BF(V) dV=4πA3V2e-BV2dV

 

 

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