An ideal gas occupies a volume of $$2 m^3$$ at a pressure of $$3 \times 10^6 Pa$$. The energy of the gas is:

The problem speaks about an ideal gas, so we recall the relationship between the internal energy $$U$$ of an ideal, mono-atomic gas and its state variables. For such a gas, the internal energy depends only on temperature and is given by the well-known formula

$$U \;=\; \frac{3}{2}\,nRT.$$

Here $$nRT$$ can be replaced by $$PV$$ through the ideal-gas equation $$PV=nRT$$. Substituting $$nRT=PV$$ into the expression for $$U$$, we arrive at

$$U \;=\; \frac{3}{2}\,PV.$$

Now we insert the numerical values supplied in the question. The pressure is

$$P \;=\; 3 \times 10^{6}\ \text{Pa},$$

and the volume is

$$V \;=\; 2\ \text{m}^{3}.$$

First we form the product $$PV$$:

$$PV \;=\; \bigl(3 \times 10^{6}\ \text{Pa}\bigr)\,\bigl(2\ \text{m}^{3}\bigr) \;=\; 6 \times 10^{6}\ \text{Pa·m}^{3}.$$

The unit $$\text{Pa·m}^{3}$$ is equivalent to the joule, because $$1\ \text{Pa}=1\ \text{N\,m}^{-2}$$ and $$1\ \text{N\,m}=1\ \text{J}$$. Hence

$$PV \;=\; 6 \times 10^{6}\ \text{J}.$$

Next we multiply by $$\dfrac{3}{2}$$ as demanded by the formula for internal energy:

$$U \;=\; \frac{3}{2}\,\bigl(6 \times 10^{6}\ \text{J}\bigr) \;=\; 3 \times 10^{6}\ \text{J} \times 1.5 \;=\; 9 \times 10^{6}\ \text{J}.$$

Thus the energy of the gas is $$9 \times 10^{6}\ \text{J}.$$

Hence, the correct answer is Option B.