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6.4: The Probability-density Function for Gas Velocities in One Dimension

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  • In Section 4.3, we find a differential equation in the function \({\rho }_x\left(v_x\right)\). Unlike the velocity, which takes values from zero to infinity, the \(x\)-component, \(v_x\), takes values from minus infinity to plus infinity. The probability density at an infinite velocity, in either direction, is necessarily zero. Therefore, we cannot evaluate the integral of \(d \rho_x\left(v_x\right)/ \rho_x\left(v_x\right)\) from \(v_x=-\infty\) to an arbitrary velocity, \(v_x\). However, we know from Maxwell’s assumption that the probability density for \(v_x\) must be independent of whether the molecule is traveling in the direction of the positive \(x\)-axis or the negative \(x\)-axis. That is, \(\rho_x\left(v_x\right)\) must be an even function; the probability density function must be symmetric around \(v_x=0\); \(\rho_x\left(v_x\right)=\rho_x\left(-v_x\right)\). Hence, we can express \(\rho_x\left(v_x\right)\) relative to its fixed value,\( \rho_x\left(0\right)\), at \(v_x=0\). We integrate \(d \rho_x\left(v_x\right)/ \rho_x\left(v_x\right)\) from \(\rho_x\left(0\right)\) to \(\rho_x\left(v_x\right)\) as \(v_x\) goes from zero to an arbitrary velocity, \(v_x\), to find

    \[\int^{\rho_x\left(v_x\right)}_{\rho_x\left(0\right)} \frac{d \rho_x\left(v_x\right)}{\rho_x\left(v_x\right)}=-\lambda \int^{v_x}_0 v_xdv_x\]


    \[\rho_x\left(v_x\right)=\frac{df_x\left(v_x\right)}{dv_x}= \rho_x\left(0\right)\mathrm{exp}\left(\frac{-\lambda v^2_x}{2}\right)\]

    The value of \(\rho_x\left(0\right)\) must be such as to make the integral of \(\rho _x\left(v_x\right)\) over all possible values of \(v_x\), \(-\infty < v_x <\infty \), equal to unity. That is, we must have

    \[\begin{aligned} 1 & =\int^{\infty}_{-\infty} \rho_x\left(v_x\right) dv_x \\ ~ & =\int^{\infty}_{-\infty} \frac{df_x\left(v_x\right)}{dv_x}dv_x \\ ~ & = \rho_x \left(0\right)\int^{\infty}_{-\infty} \mathrm{exp} \left(\frac{-\lambda v^2_x}{2}\right) dv_x \\ ~ & =\rho_x\left(0\right)\sqrt{\frac{2\pi }{\lambda }} \end{aligned}\]

    where we use the definite integral \(\int^{\infty }_{-\infty } \mathrm{exp} \left(-ax^2\right) dx=\sqrt{ \pi /a}\). (See Appendix D.) It follows that \( \rho_x\left(0\right)= \left( \lambda /2 \pi \right)^{1/2}\). The one-dimensional probability-density function becomes

    \[\begin{aligned} \rho_x\left(v_x\right) & =\frac{df_x\left(v_x\right)}{dv_x} \\ ~ & = \left(\frac{\lambda }{2\pi }\right)^{1/2}\mathrm{exp}\left(\frac{-\lambda v^2_x}{2}\right) \end{aligned}\]

    Note that this is the normal distribution with \(\mu =0\) and \({\sigma }^2={\lambda }^{-1}\). So \({\lambda }^{-1}\) is the variance of the normal one-dimensional probability-density function. As noted above, in Section 4.6 we find that \(\lambda ={m}/{kT}\).