Question

Question: In a gas that contains \(N\) molecules, is it accurate to say that the number of molecules with speed \(v\)is equal to \(f\left( v \right)\)? Is it accurate to say that this number is given by \(Nf\left( v \right)\)? Explain your answers.

Short answer

The \(Nf\left( v \right)\), is not the number of the molecules with the exactly speed \(v\). But the number of the molecules is given by \(Nf\left( v \right)\).

Step-by-step solution

Write the given data from the question.

The number of molecules of the gas is \(N\).

The speed of the molecules is \(v\).

Determine the formulas to calculate the number of the molecules.

The expression to calculate the number of the molecules with the speed is given as follows.

\(\Delta N = N\int_v^{\Delta v + v} {f\left( v \right)dv} \) …… (i)

Here,\(f\left( v \right)\)is the Maxwell-Boltzmann distribution.

Calculate the number of the molecules.

Let’s assume \(f\left( v \right)\) is very small.

Rewrite the equation (i).

\(\Delta N = Nf\left( v \right)\Delta v\)

Substitute \(1\) for \(\Delta v\) into above equation.

\(\begin{array}{l}\Delta N = Nf\left( v \right)\left( 1 \right)\\\Delta N = Nf\left( v \right)\end{array}\)

From the above discussion, \(Nf\left( v \right)\) is the number of the molecules and the \(Nf\left( v \right)\) is the number of the molecules with the speed interval \(v\) to \(v + \Delta v\) and \(f\left( v \right)\) represents the probability per unit speed interval. Therefore, \(Nf\left( v \right)\), is not the number of the molecules with the exactly speed \(v\).

Hence the \(Nf\left( v \right)\), is not the number of the molecules with the exactly speed \(v\). But the number of the molecules is given by \(Nf\left( v \right)\).