Question

$n$ moles of helium gas are placed in a vessel of volume $V$ liter at $T K$. 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. $\left[\frac{6 V_{1}}{\pi N_{ A } n}\right]^{1 / 3}$

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} $

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

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