An ideal gas consists of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{mol}\mathbf{}$of diatomic molecules that rotate but do not oscillate. The molecular diameter is $\mathbf{250}\mathbf{pm}$. The gas is expanded at a constant pressure of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^5\mathbf{}\mathbf{Pa}$,with a transfer of $\mathbf{200}\mathbf{J}\mathbf{}$as heat. what is the change in the mean free path of the molecules?

To find:

Change in the mean free path of the molecules.

We know the formula for mean free path. Here, we are giventhe heat energy and the pressure. The process is at constant pressure. Using the ideal gas equation and formula for heat generated, we rearrange the equation for mean free path in terms of the given values, and from this, we get the required answer.

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

$$\begin{gathered} \text{PV}=\text{nR}\Delta \text{T} \\ \text{Q}={\text{nC}}_p\Delta \text{T} \end{gathered}$$

$\text{P}=1.5\times {10}^5\text{Pa}$

$$\begin{gathered} \text{d}=250\text{pm}=250\times {10}^{-12}\text{m} \\ \text{Q}=200\text{\hspace{0.17em}J} \end{gathered}$$

We have

$\lambda =\frac{1}{\sqrt{2}\pi {\text{d}}^2\left(\frac{\text{N}}{\text{V}}\right)}$

But,

$$\begin{gathered} \text{PV}=\text{nR}\Delta \text{T} \\ \text{V}=\frac{\text{nR}\Delta \text{T}}{\text{P}} \\ \lambda =\frac{\text{nR}\Delta \text{T}}{\sqrt{2}\pi {\text{d}}^2\text{PN}} \end{gathered}$$

We also have

$$\begin{gathered} \text{Q}={\text{nC}}_p\Delta \text{T} \\ \text{n}=\frac{\text{Q}}{{\text{C}}_p\Delta \text{T}} \\ \Delta \lambda =\frac{\text{R}\Delta \text{TQ}}{\sqrt{2}\pi {\text{d}}^2{\text{PNC}}_p\Delta \text{T}} \\ \Delta \lambda =\frac{\text{RQ}}{\sqrt{2}\pi {\text{d}}^2{\text{PNC}}_p} \end{gathered}$$

Where

$$\begin{gathered} \text{N}=\left(\text{Numberofmoles}\right)\times 6.022\times {10}^{23} \\ \text{N}=1.5\times 6.022\times {10}^{23} \\ \text{N}=9\times {10}^{23} \end{gathered}$$

For constant pressure

$$\begin{gathered} {\text{C}}_p=\left(\frac{7}{2}\right)\text{R} \\ \Delta \lambda =\frac{\text{RQ}}{\sqrt{2}\pi {\text{d}}^2\text{PN}\times \left(\frac{7}{2}\right)\text{R}} \\ \Delta \lambda =\frac{2\text{Q}}{7\sqrt{2}\pi {\text{d}}^2\text{PN}} \\ \Delta \lambda =\frac{2\times 200}{7\sqrt{2}\pi \times {(250\times {10}^{-12})}^2\times (1.5\times {10}^5)\times (9\times {10}^{23})} \\ \Delta \lambda =1.52\times {10}^{-9}\text{m}=1.52\text{\hspace{0.17em}nm} \end{gathered}$$

Final Statement:

The change in mean free path of molecules is found to be$1.52\text{\hspace{0.17em}nm}.$