NEET 2020 Physics Question 139 Solution

Starting with the ideal gas law: $$P V = n R T = \left(\frac{m}{M}\right) R T$$ Rearranging for density $$\rho = \frac{m}{V}$$: $$\rho = \frac{P M}{R T}$$

- Absolute Temperature: $$T = 27^\circ\text{C} + 273 = 300\text{ K}$$ - Pressure: $$P = 249\text{ kPa} = 249 \times 10^3\text{ Pa}$$ - Universal Gas Constant: $$R = 8.3\text{ J mol}^{-1}\text{K}^{-1}$$ - Molar mass of hydrogen gas ($$\text{H}_2$$): $$M = 2\text{ g mol}^{-1} = 2 \times 10^{-3}\text{ kg mol}^{-1}$$

$$\rho = \frac{(249 \times 10^3\text{ N m}^{-2}) \times (2 \times 10^{-3}\text{ kg mol}^{-1})}{(8.3\text{ J mol}^{-1}\text{K}^{-1}) \times (300\text{ K})}$$ The factors of $$10^3$$ and $$10^{-3}$$ in the numerator cancel: $$\text{Numerator} = 249 \times 2 = 498\text{ kg m}^{-2}$$ $$\text{Denominator} = 8.3 \times 300 = 2490\text{ J K}^{-1} = 2490\text{ N m}$$ $$\rho = \frac{498}{2490} = \frac{1}{5} = 0.2\text{ kg m}^{-3}$$

The calculated density is exactly $$0.2\text{ kg/m}^3$$, matching Option (2).