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Question: In moles of Helium gas are placed in a vessel of volume V litre at T K. If V<sub>1</sub> is ideal vo...

In moles of Helium gas are placed in a vessel of volume V litre at T K. If V1 is ideal volume of Helium then diameter of He –atom is –

A

32(V1πNAn)1/3\frac{3}{2}\left( \frac{V_{1}}{\pi N_{A}n} \right)^{1/3}

B

32[(V–V1)πNAn]1/3\frac{3}{2}\left\lbrack \frac{\left( V–V_{1} \right)}{\pi N_{A}n} \right\rbrack^{1/3}

C

[6(V–V1)πNAn]1/3\left\lbrack \frac{6(V–V_{1})}{\pi N_{A}n} \right\rbrack^{1/3}

D

[6V1πNAn]1/3\left\lbrack \frac{6V_{1}}{\pi N_{A}n} \right\rbrack^{1/3}

Answer

[6(V–V1)πNAn]1/3\left\lbrack \frac{6(V–V_{1})}{\pi N_{A}n} \right\rbrack^{1/3}

Explanation

Solution

Volume of n moles of He atoms = V–V1

Ž Volume of 1 atoms of He = (V–V1)NAn\frac{(V–V_{1})}{N_{A}n}

Ž 43\frac{4}{3} pr3 = V–V1NAn\frac{V–V_{1}}{N_{A}n}

Ž r = {34(V–V1NAn)}1/3\left\{ \frac{3}{4}\left( \frac{V–V_{1}}{N_{A}n} \right) \right\}^{1/3}

Ž 2r = {23×34(V–V1NAn)}1/3\left\{ \frac{2^{3} \times 3}{4}\left( \frac{V–V_{1}}{N_{A}n} \right) \right\}^{1/3}

Ž So diameter = {6(V–V1)πNAn}1/3\left\{ \frac{6(V–V_{1})}{\pi N_{A}n} \right\}^{1/3}