Question

The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy $2 \times 10^{-9}$ J per molecule is :

• A. $0.75 \times 10^{11}$
• B. $1.5 \times 10^{11}$
• C. $3 \times 10^{11}$
• D. $6 \times 10^{11}$

Hint:

In kinetic theory problems, use $P=\tfrac{2}{3}(U/V)$ directly when mean kinetic energy per molecule is given. Always convert pressure and volume into SI units first.

Answer 1

The Correct Option is B

From kinetic theory of gases, the pressure of an ideal gas is related to the total kinetic energy $U$ by: \[ P = \frac{2}{3}\,\frac{U}{V} \] The total kinetic energy is: \[ U = N \times \overline{K} \] where $\overline{K}$ is the mean kinetic energy per molecule. Substituting, \[ P = \frac{2}{3}\,\frac{N\overline{K}}{V} \] Solving for $N$, \[ N = \frac{3PV}{2\overline{K}} \] Substitute given values (SI units): \[ P = 2~\text{atm} = 2\times10^5~\text{Pa} \] \[ V = 1~\text{L} = 10^{-3}~\text{m}^3 \] \[ \overline{K} = 2\times10^{-9}~\text{J} \] \[ N = \frac{3(2\times10^5)(10^{-3})}{2(2\times10^{-9})} \] \[ N = \frac{6\times10^2}{4\times10^{-9}} = 1.5\times10^{11} \] \[ \boxed{N = 1.5\times10^{11}} \]