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:

We are given a sample of ideal gas whose volume is one litre. Remember that one litre must first be turned into SI units:

$$1 \text{ litre}=1 \times 10^{-3}\ \text{m}^3$$

The pressure is quoted as $$2$$ atmospheres. To write this in pascals, we recall that

$$1\ \text{atm}=1.013\times 10^{5}\ \text{Pa}.$$

So,

$$P=2\ \text{atm}=2 \times 1.013\times 10^{5}\ \text{Pa}=2.026\times 10^{5}\ \text{Pa}.$$

For an ideal gas, the kinetic theory relation connecting pressure, volume and the average translational kinetic energy per molecule is first stated:

$$PV=\frac{2}{3} N \, \langle E_{\text{kin}}\rangle,$$

where

$$N$$ is the required number of molecules and

$$\langle E_{\text{kin}}\rangle$$ is the mean kinetic energy per molecule.

The problem itself tells us that each molecule has an average kinetic energy

$$\langle E_{\text{kin}}\rangle = 2 \times 10^{-9}\ \text{J}.$$

Now we substitute the known values of $$P$$ and $$V$$ into $$PV$$:

$$PV = \left(2.026\times 10^{5}\ \text{Pa}\right)\left(1\times 10^{-3}\ \text{m}^3\right).$$

Multiplying we obtain

$$PV = 2.026\times 10^{2}\ \text{J}=202.6\ \text{J}.$$

Next we insert this result and the given kinetic energy into the kinetic‐theory equation:

$$202.6\ \text{J} = \frac{2}{3} N \left(2\times 10^{-9}\ \text{J}\right).$$

To isolate $$N$$, we first multiply both sides by $$3$$:

$$3 \times 202.6\ \text{J} = 2 N \left(2\times 10^{-9}\ \text{J}\right).$$

This simplifies to

$$607.8\ \text{J} = 4 \times 10^{-9}\ \text{J}\times N.$$

Finally, we divide both sides by $$4 \times 10^{-9}\ \text{J}$$ to solve for $$N$$:

$$N=\frac{607.8}{4 \times 10^{-9}}.$$

Carrying out the division,

$$N = 1.5195 \times 10^{11}.$$

Rounding appropriately,

$$N \approx 1.5 \times 10^{11}.$$

Hence, the correct answer is Option C.