Question

A container contains 1 mol of nitrogen gas at temperature 17${}^o$C and pressure 2 atmospheres. If the radius of nitrogen molecule is 1 A${}^o$ then the volume of gas will be

• A. 46.52 litre
• B. 23.76 litre
• C. 1.19 litre
• D. 5.42 litre

Answer 1

The correct option is C 1.19 litre

Given : $T=17+273=290$ K $P=2atm=2\times 101325$ Pa and $n=1$

The real gas equation predicted by the Van Der Waal's law is given by

$(P+\frac{n^2}{V^2}a)(V-nb)=nRT$

Under high pressures and at lower temperatures, the term $(V-nb)$ majorly contributes to the deviation from Ideal gases.

Thus, $V\approx nb+\frac{nRT}{P}$

where $b$ is the volume occupied by one mole of the gas molecules.

Give that radius of the nitrogen molecule is $1{\text{A}}^{\circ }=1\times {10}^{-10}\text{m}$

Thus, volume of one molecule of nitrogen is $\frac{4}{3}\pi a^3=\frac{4}{3}\pi \times (1\times {10}^{-10}{)}^3=4.1887\times {10}^{-30}{\text{m}}^3$

Thus, volume of one mole of nitrogen is $b=6.022\times {10}^{23}\times 4.1887\times {10}^{-30}=25.22\times {10}^{-7}{\text{m}}^3/\text{mol}$

Thus, volume of gas $V=25.22\times {10}^{-7}+\frac{8.314\times 290}{2\times 101325}=(11.897+0.002522)\times {10}^{-3}{m}^3=11.9\text{litre}$