The molar volume of  He at 10.1325 MPa and 273K  is 0.011075 of its molar volume at 101.325 kPa at 273 K. Calculate the radius of helium atom. The gas is assumed to show real gas nature. Neglect the value of  $a$  for He. 

For real gas :  $[P+\frac{a}{V^2}][V-b]=RT$
OR  $P[V-b]=RT\left(neglectinga\right)$
 $\therefore \frac{10.1325\times {10}^6}{101325}[V_1-b]=0.0821\times 273(101325Pa=1atm)$
OR  $100[V_1-b]=0.0821\times 273=22.41\mathrm{.........}\left(i\right)$
OR  $\frac{101.325\times {10}^3}{101325}[V_2-b]=0.0821\times 273$
OR   $[V_2-b]=22.41\mathrm{..........}\left(ii\right)$
By Eq. (i),  $V_1=0.2241+b$               ……… (iii) 
By Eq. (ii),    $V_2=22.41+b$              ……… (iv)   
By Eqs. (iii) and (iv),  $\frac{V_1}{V_2}=\frac{0.2241+b}{22.41+b}$
 $\frac{0.011075V_2}{V_2}=\frac{0.2241+b}{22.41+b}$
  $(V_1=0.011075V_{2}isgiven)$
$\therefore b=0.024$  litre  $mo{l}^{-1}=24c{m}^{3}mo{l}^{-1}$
 $\because b=4N\times v=4\times 6.023\times {10}^{23}\times \frac{4}{3}\pi r^3$
Or  $24=4\times 6.023\times {10}^{23}\times \frac{4}{3}\times \frac{22}{7}\times r^3$
 $\therefore r=1.33\times {10}^{-8}cm$