2 kg of a monoatomic gas is at a pressure of $$4 \times 10^4$$ N m$$^{-2}$$. The density of the gas is 8 kg m$$^{-3}$$. What is the order of energy of the gas due to its thermal motion?

We have a mono-atomic ideal gas whose mass is given as $$m = 2\ \text{kg}$$. The pressure of the gas is given as $$P = 4 \times 10^{4}\ \text{N m}^{-2}$$ and its density is $$\rho = 8\ \text{kg m}^{-3}$$.

Density is related to mass and volume by the formula $$\rho = \dfrac{m}{V}.$$

Re-arranging, the volume of the gas is obtained as $$ V = \dfrac{m}{\rho}. $$

Substituting the given values, $$ V = \dfrac{2\ \text{kg}}{8\ \text{kg m}^{-3}} = 0.25\ \text{m}^{3}. $$

For an ideal gas, the product of pressure and volume equals $$nRT$$, i.e. $$ PV = nRT. $$ Although the actual value of $$nRT$$ is not required, we still use the equality $$PV = nRT$$ to replace $$nRT$$ in the expression for internal energy.

The total thermal (internal) energy of a mono-atomic ideal gas is given by the formula $$ U = \dfrac{3}{2}\,nRT. $$ Using $$PV = nRT,$$ we can rewrite this as $$ U = \dfrac{3}{2}\,PV. $$

Now we first evaluate $$PV$$: $$ PV = \left(4 \times 10^{4}\ \text{N m}^{-2}\right)\left(0.25\ \text{m}^{3}\right) = 1.0 \times 10^{4}\ \text{J}. $$ (The unit N m−2 × m3 simplifies to joules.)

Substituting this value of $$PV$$ into the internal-energy formula, $$ U = \dfrac{3}{2}\,PV = \dfrac{3}{2}\left(1.0 \times 10^{4}\ \text{J}\right) = 1.5 \times 10^{4}\ \text{J}. $$

The numerical value $$1.5 \times 10^{4}\ \text{J}$$ lies in the order of $$10^{4}\ \text{J}.$$

Hence, the correct answer is Option C.