An open glass tube is immersed in mercury in such a way that a length of 8cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm. What will be the length of air column above mercury in the tube now? (atm pressure= 76 cm of Hg)
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Arrangement of glass tube as shown in figure.
Because temperature doesn't change . So, apply Boyle's law
P₁V₁ = P₂V₂
Here P is the pressure and V is the volume of tube .
∵ V = A × h [ A denotes base area and h denotes height ]
no change in base area then Boyle's law is converted
P₁ × h₁ = P₂ × h₂
Here P₁ = Patm = 76cm Hg , h₁ = 8cm , h₂ = (54 - x) cm
∴ 76 × 8 = P₂( 54 - x) ------(1)
Again, we see that, P₂ + x = Patm = 76cm
so, P₂ = (76 - x) -----(2), put it in equation (1)
76 × 8 = (76 - x)(54 - x)
Solve it x = 38 and 92
But x ≠ 92 cm because x ≤ 54cm
So, length of air column above Mercury = (54 - x) = 16 cm
Because temperature doesn't change . So, apply Boyle's law
P₁V₁ = P₂V₂
Here P is the pressure and V is the volume of tube .
∵ V = A × h [ A denotes base area and h denotes height ]
no change in base area then Boyle's law is converted
P₁ × h₁ = P₂ × h₂
Here P₁ = Patm = 76cm Hg , h₁ = 8cm , h₂ = (54 - x) cm
∴ 76 × 8 = P₂( 54 - x) ------(1)
Again, we see that, P₂ + x = Patm = 76cm
so, P₂ = (76 - x) -----(2), put it in equation (1)
76 × 8 = (76 - x)(54 - x)
Solve it x = 38 and 92
But x ≠ 92 cm because x ≤ 54cm
So, length of air column above Mercury = (54 - x) = 16 cm
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