91Ó°ÊÓ

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.

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{d}^2\left(\frac{N}{V}\right)}$

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

$\text{P}=1.5×{10}^5\text{Pa}$

$$\begin{gathered} \text{d}=250\text{pm}=250×{10}^{−12}\text{m} \\ \text{Q}=200\text{ J} \end{gathered}$$

We have

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{\text{d}}^2\left(\frac{\text{N}}{\text{V}}\right)}$

But,

$$\begin{gathered} \text{PV}=\text{nR}\mathrm{Δ}\text{T} \\ \text{V}=\frac{\text{nR}\mathrm{Δ}\text{T}}{\text{P}} \\ \mathrm{λ}=\frac{\text{nR}\mathrm{Δ}\text{T}}{\sqrt{2}\mathrm{π}{\text{d}}^2\text{PN}} \end{gathered}$$

We also have

$$\begin{gathered} \text{Q}={\text{nC}}_p\mathrm{Δ}\text{T} \\ \text{n}=\frac{\text{Q}}{{\text{C}}_p\mathrm{Δ}\text{T}} \\ \mathrm{Δ}\mathrm{λ}=\frac{\text{R}\mathrm{Δ}\text{TQ}}{\sqrt{2}\mathrm{π}{\text{d}}^2{\text{PNC}}_p\mathrm{Δ}\text{T}} \\ \mathrm{Δ}\mathrm{λ}=\frac{\text{RQ}}{\sqrt{2}\mathrm{π}{\text{d}}^2{\text{PNC}}_p} \end{gathered}$$

Where

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

For constant pressure

$$\begin{gathered} {\text{C}}_p=\left(\frac{7}{2}\right)\text{R} \\ \mathrm{Δ}\mathrm{λ}=\frac{\text{RQ}}{\sqrt{2}\mathrm{Ï€}{\text{d}}^2\text{PN}×\left(\frac{7}{2}\right)\text{R}} \\ \mathrm{Δ}\mathrm{λ}=\frac{2\text{Q}}{7\sqrt{2}\mathrm{Ï€}{\text{d}}^2\text{PN}} \\ \mathrm{Δ}\mathrm{λ}=\frac{2×200}{7\sqrt{2}\mathrm{Ï€}×{(250×{10}^{−12})}^2×(1.5×{10}^5)×(9×{10}^{23})} \\ \mathrm{Δ}\mathrm{λ}=1.52×{10}^{−9}\text{m}=1.52\text{ n³¾} \end{gathered}$$

Final Statement:

The change in mean free path of molecules is found to be$1.52\text{ n³¾}.$