91Ó°ÊÓ

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:

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

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:

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

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:

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

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