As a sample of nitrogen gas (N₂) undergoes a temperature increase at constant volume, the distribution of molecular speeds increases. That is, the probability distribution function P(v)for the molecules spreads to higher speed values, as suggested in Fig. 19-8b. One way to report the spread in P(v)is to measure the difference$\mathbf{∆}\mathit{v}$ between the most probable speed${\mathit{v}}_{\mathbf{p}}$and the rms speed${\mathit{v}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$. When P(v) spreads to higher speeds,$\mathbf{∆}\mathit{v}$ increases. Assume that the gas is ideal and the N₂ molecules rotate but do not oscillate. For 1.5 mol, an initial temperature of 250 K, and a final temperature of 500 K, what are (a) the initial difference$\mathbf{∆}{\mathit{v}}_{\mathbf{i}}$, (b) the final difference,$\mathbf{∆}{\mathit{v}}_{\mathbf{f}}$, and (c) the entropy change$\mathbf{∆}\mathit{S}$ for the gas?

The number of moles of an ideal gas, n = 1.5 mol

The initial temperature,${\mathrm{T}}_{\mathrm{i}}=250\mathrm{K}$

The final temperature,${\mathrm{T}}_f=500\mathrm{K}$

The molar mass of Nitrogen gas (N₂),$\mathrm{M}=0.028\mathrm{kg}/\mathrm{mol}$

We need to use the equation for change in entropy related to the final temperature and initial temperature. For the change in speed, we can use the equations from a speed and most probable speed.

Formulae:

The root mean square velocity of the gas, $v_{rms}=\sqrt{\frac{3RT}{M}}$ (1)

The most probable speed of a gas, $v_p=\sqrt{\frac{2RT}{M}}$ (2)

The entropy change of a gas, $∆\mathrm{S}=\mathrm{nRln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)+{\mathrm{nC}}_{\mathrm{v}}\ln \left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$ (3)

For the initial difference of the rms speed and probable speed, we have to use the value of initial temperature in equation (1) and equation (2). Hence, it is given as:

$$\begin{gathered} ∆{\mathrm{v}}_{\mathrm{i}}=\sqrt{\frac{3{\mathrm{RT}}_{\mathrm{i}}}{\mathrm{M}}}-\sqrt{\frac{2{\mathrm{RT}}_{\mathrm{i}}}{\mathrm{M}}} \\ =\sqrt{\frac{3\times 8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 250\mathrm{K}}{0.028\mathrm{kg}}}-\sqrt{\frac{2\times 8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 250\mathrm{K}}{0.028\mathrm{kg}}} \\ =86.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the initial difference of speeds is 86.6 m/s

For the final difference of the rms speed and probable speed, we have to use the value of initial temperature in equation (1) and equation (2). Hence, it is given as:

$$\begin{gathered} ∆{\mathrm{v}}_{\mathrm{i}}=\sqrt{\frac{3{\mathrm{RT}}_f}{\mathrm{M}}}-\sqrt{\frac{2{\mathrm{RT}}_f}{\mathrm{M}}} \\ =\sqrt{\frac{3\times 8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 500\mathrm{K}}{0.028\mathrm{kg}}}-\sqrt{\frac{2\times 8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\times 500\mathrm{K}}{0.028\mathrm{kg}}} \\ =122\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the final difference of the speeds is 122 m/s

For constant volume,$V_f-V_i$. So, the entropy change using equation (3) can be given as:

$$\begin{gathered} ∆S=\frac{5}{2}nR\ln \left(\frac{T_f}{T_i}\right) \\ =\frac{5}{2}\times 1.5\times 8.314J/mol·K\times \ln \left(\frac{500K}{250K}\right) \\ =22J/K \end{gathered}$$

Hence, the value of the entropy change is 22 J/K