Question

The molar volume of helium at $10000atm$ and $300K$ is $0.01$ times to the molar volume of $He$ at $1atm$ and $300K$. The value of $a$ for $He$ can be neglected. Determine the radius of $He$ atom in pm in nearest possible integers.

Answer 1

Answer: 290

$P\left(V_m-b\right)=RT$
Let molar volume at 10000 atm and $300K=V_m$
$10^4\left(V_m-b\right)=0.082\times 300$
$\therefore 1\times \left(100V_m-b\right)=0.082\times 300$ ..... (I)
$\therefore 1\times \left(100V_m-b\right)=0.082\times 300$.....(II)
(I) $÷$ (II) gives.
$\frac{10^4\left(V_m-b\right)}{\left(100V_m-b\right)}=1$
or $10^4{V}_m-10^{4}b=100V_m-b$
or $\left(10^4-100\right)V_m=\left(10^4+1\right)b$
or $V_m=\frac{\left(10^4+1\right)}{9900}\times b=1.0102b$
$\therefore$ substituting the value of $b$ is equation (II)
$1\times (101.02-1)b=0.082\times 300$
or $b\times 100.02=0.082\times 300$
or $b=0.246$
or $b=0.246c{m}^3/mol$
$b=4\times N_A\times \pi r^3$
$\therefore 4\times 6\times 10^{23}\times \pi \times r^3$
or $r^3=\times 10^{-23}=\times 10^{-24}$
or $r^3=24.48\times 10^{-24}$
or $r=2.9\times 10^{-8}cm=2.9\times 10^{-10}m$.
or $r=290pm$.