1. Use the ideal gas relation in molecular form

For an ideal gas, $PV = NkT$ where:

• $P$ = pressure
• $V$ = volume
• $N$ = number of molecules
• $k$ = Boltzmann constant
• $T$ = absolute temperature

So, $N = \frac{PV}{kT}$

2. Relate mean kinetic energy to temperature

For an ideal gas, the mean translational kinetic energy per molecule is $\frac{3}{2}kT$

Given: $\frac{3}{2}kT = 2 \times 10^{-9}\,\text{J}$

Hence, $kT = \frac{2}{3}\cdot 2 \times 10^{-9} = \frac{4}{3}\times 10^{-9}\,\text{J}$

3. Substitute given pressure and volume

Given:

• $P = 2$ atmospheric pressure
• $V = 1$ litre

Using standard values: $1\,\text{atm} \approx 10^5\,\text{Pa}, \qquad 1\,\text{litre} = 10^{-3}\,\text{m}^3$

Therefore, $P = 2 \times 10^5\,\text{Pa}$ $V = 10^{-3}\,\text{m}^3$

So, $PV = 2 \times 10^5 \times 10^{-3} = 200\,\text{J}$

4. Calculate number of molecules

$N = \frac{PV}{kT} = \frac{200}{\frac{4}{3}\times 10^{-9}}$

$N = 200 \times \frac{3}{4} \times 10^9$

$N = 150 \times 10^9 = 1.5 \times 10^{11}$

5. Match with options

$1.5 \times 10^{11}$ corresponds to Option C.

Note

The stated temperature $300\,\text{K}$ is inconsistent with the given mean kinetic energy $2\times10^{-9}\,\text{J}$ per molecule, since at $300\,\text{K}$ the mean kinetic energy should be of order $10^{-21}\,\text{J}$. The question is evidently intended to be solved using the given kinetic energy value together with $PV=NkT$.