MMV²dN= 47(-N)3/2y² exp(-2 RTF(V)dV =-)dv-)dV2RTFind the Vm value by considering the following curve.ax 1020 x 10298 K15 x 1010 x 101500 K5x 10Vm Vm|10002

Answered: MV². 2RT dN M F(V)dV =5 N = 47(- 3:2y²…

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Question

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 $V_{m}$ 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

$F (V) d V = \frac{d N}{N} = 4 π {(\frac{M}{2 πRT})}^{3 / 2} V^{2} \exp (- \frac{{MV}^{2}}{2 RT}) dV From the concept of maxima and minima, at V = V_{m} the derivative of the distribution function will be 0 substituting \sqrt{\frac{M}{2 πRT}} = A and \frac{M}{2 RT} = B F (V) dV = 4 {πA}^{3} V^{2} e^{- B V^{2}} dV$

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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