Question

$n$ moles of helium gas are placed in a vessel of volume $V$ liter at $TK$. If $V_1$ is ideal volume of Helium, then diameter of $He$ - atom is

• A. $\frac{3}{2}{\left(\frac{V_1}{\pi N_{A}n}\right)}^{1/3}$
• B. $\frac{3}{2}{\left[\frac{\left(V-V_1\right)}{\pi N_{A}n}\right]}^{1/3}$
• C. ${\left[\frac{6\left(V-V_1\right)}{\pi N_{A}n}\right]}^{1/3}$
• D. ${\left[\frac{6V_1}{\pi N_{A}n}\right]}^{1/3}$

Answer 1

Answer: (C)

Volume of $n$ moles of $He$ atoms $=V-V_1$

$\Rightarrow$ Volume of 1 atom of $He=\frac{\left(V-V_1\right)}{N_{A}n}$

$\Rightarrow \frac{4}{3}\pi r^3=\frac{V-V_1}{N_{A}n}$

$\Rightarrow r={\left\{\frac{3}{4\pi }=\left(\frac{V-V_1}{N_{A}n}\right)\right\}}^{1/3}$

$\Rightarrow 2r={\left\{\frac{2^3\times 3}{4\pi }\left(\frac{V-V_1}{N_{A}n}\right)\right\}}^{1/3}$

$\Rightarrow$ So diameter $={\left\{\frac{6\left(V-V_1\right)}{\pi N_{A}n}\right\}}^{1/3}$