Question

Assuming the molecules of gas as hard sphere of radius $2.0\times {10}^{-10}$m the fraction of volume occupied by the molecules to the total volume of a given amount of gas at ${27}^{o}C$ and at 1 bar pressure and 10 bar pressure respectively are

• A. 99.9%, 99%
• B. 0.082%, 0.82%
• C. 99%, 90%
• D. 1%, 10%

Answer 1

The correct option is B 0.082%, 0.82%

At $P=1bar$
Volume of gas $=\frac{nRT}{P}=\frac{n\times 0.0821\times 300}{1}=24.63\times nL$
At $P=10bar$
Volume of gas $=\frac{n\times 0.0821\times 300}{100}=2.463\times nL$
Also, volume of molecules $=n\times Avogadr{o}^{\prime }sno.\times \frac{4}{3}\pi r^3$
$=n\times 6.023\times {10}^{23}\times \frac{4}{3}\times 3.14\times (2.0\times {10}^8{)}^3$
$=20.17\times nc{m}^3$
$=0.0202\times nL$
At $P=1$ bar
$\therefore \frac{V_m}{V_g}=\frac{0.0202\times n}{24.63\times n}=8.2\times {10}^{-4}$
or $8.2\times {10}^{-2}$% =0.082%
At $P=10bar$
$\frac{V_m}{V_g}=\frac{0.0202}{2.463\times n}=8.2\times {10}^{-3}$
or $8.2\times {10}^{-1}$% $=$ 0.82%