There is a steel cylinder with oxygen of mass m and density ρ. When oxygen of mass m / 4 is used, the remaining gas density in the cylinder is 0

There is a steel cylinder with oxygen of mass m and density ρ. When oxygen of mass m / 4 is used, the remaining gas density in the cylinder is 0

Cylinder volume v = m / ρ
The volume of gas in the cylinder is always equal to the volume m / ρ of the cylinder
After using M / 4 oxygen, the mass of oxygen in the bottle is 3m / 4
Density = (3m / 4) / (M / ρ) = 3 ρ / 4

The density of oxygen in a bottle is 6kg / m3. If 13 of them are used at one time, the density of residual oxygen in the bottle is 6kg / m3______ ．

In the whole process, the volume of oxygen remains unchanged. If the volume of the oxygen cylinder is V, then the original mass of oxygen is m = ρ v. after 13 is used, the remaining m ′ = (1-13) ρ v = 23 ρ V, then the remaining oxygen density in the cylinder is: ρ′ = m ′ = 23 ρ VV = 23 × 6kg / m3 = 4kg / m3, so the answer is: 4kg / m3

The density of oxygen in a steel cylinder is 8kg / m3. If 14% of its mass is used in gas welding on a certain day, the density of residual oxygen in the cylinder is ()
A. 8kg/m3B. 6kg/m3C. 4kg/m3D. 2kg/m3

Suppose the volume of the cylinder is V, the mass of oxygen in the original cylinder is: M0 = ρ 0V, 14 of which is used, the mass of residual oxygen is: M = 34m0 = 34 ρ 0V, ∵ the volume of oxygen in the cylinder remains unchanged, ∵ the density of residual oxygen is: ρ = MV = 34 ρ 0vv = 34 ρ 0 = 34 × 8kg / m3 = 6kg / m3