Back to Directory
NEET PhysicsMedium

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

1

0.5 kg/m3^3

2

0.2 kg/m3^3

3

0.1 kg/m3^3

4

0.02 kg/m3^3

Step-by-Step Solution

Using PM=ρRTPM = \rho RT, we get ρ=PMRT\rho = \frac{PM}{RT}. Given P=249×103P = 249 \times 10^3 N/m2^2, M=2×10−3M = 2 \times 10^{-3} kg, T=300T = 300 K. Substituting these, ρ=(249×103)(2×10−3)8.3×300=0.2\rho = \frac{(249 \times 10^3)(2 \times 10^{-3})}{8.3 \times 300} = 0.2 kg/m3^3.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started