Question

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

Answer 1

$P\left(V_{m}-b\right)=R T$
Let molar volume at 10000 atm and $300\, K =V_{m}$
$10^{4}\left(V_{m}-b\right)=0.082 \times 300$
$\therefore 1 \times\left(100 V_{m}-b\right)=0.082 \times 300$ ..... (I)
$\therefore 1 \times\left(100 V_{m}-b\right)=0.082 \times 300$.....(II)
(I) $\div$ (II) gives.
$\frac{10^{4}\left(V_{m}-b\right)}{\left(100 \,V_{m}-b\right)}=1$
or $ 10^{4} V_{m}-10^{4} b=100 \,V_{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=1.0102\, b$
$\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 $b=0.246 \,cm ^{3} / mol$
$b=4 \times N_{A} \times \pi r^{3}$
$\therefore 4 \times 6 \times 10^{23} \times \pi \times r^{3}$
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 $r=290\, pm$.