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NEET PHYSICSEasy

A cylinder contains hydrogen gas at a pressure of 249 kPa249 \text{ kPa} and temperature 27∘C27^\circ\text{C}. Its density is: (R=8.3 J mol−1K−1R=8.3 \text{ J mol}^{-1}\text{K}^{-1})

A

0.2 kg/m30.2 \text{ kg/m}^3

B

0.1 kg/m30.1 \text{ kg/m}^3

C

0.02 kg/m30.02 \text{ kg/m}^3

D

0.5 kg/m30.5 \text{ kg/m}^3

Step-by-Step Solution

  1. Identify the Formula: The density (ρ\rho) of an ideal gas is given by the relation ρ=PMRT\rho = \frac{PM}{RT}, derived from the ideal gas equation PV=nRTPV = nRT.
  2. Convert Units to SI: Pressure (PP) = 249 kPa=249×103 Pa249 \text{ kPa} = 249 \times 10^3 \text{ Pa}. Temperature (TT) = 27∘C=27+273=300 K27^\circ\text{C} = 27 + 273 = 300 \text{ K}.
  • Molar Mass (MM) of Hydrogen (H2H_2) = 2 g/mol=2×10−3 kg/mol2 \text{ g/mol} = 2 \times 10^{-3} \text{ kg/mol}.
  1. Substitute Values: ρ=(249×103)×(2×10−3)8.3×300\rho = \frac{(249 \times 10^3) \times (2 \times 10^{-3})}{8.3 \times 300}
  2. Calculate: Denominator: 8.3×300=24908.3 \times 300 = 2490. Numerator: 249×2=498249 \times 2 = 498. ρ=4982490=0.2 kg/m3\rho = \frac{498}{2490} = 0.2 \text{ kg/m}^3.
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