A glass flask is filled up to a mark with 50 c c of mercury at 18° C. If the flask and contents are heated to 38° C, how much mercury will be above the mark? ( α for glass is 9 × 10-6, ° C and coefficient of real expansion of mercury is 180 × 10-6 / ° C )

Question

A glass flask is filled up to a mark with 50 c c of mercury at 18° C. If the flask and contents are heated to 38° C, how much mercury will be above the mark? ( α for glass is 9 × 10-6, ° C and coefficient of real expansion of mercury is 180 × 10-6 / ° C ) (A) 0.85 c c (B) 0.46 c c (C) 0.153 c c (D) 0.05 c c

Tardigrade Admin · Accepted Answer

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by

Δ V = V_{0} [γ_{L} - γ_{g}] Δ θ = V [γ_{L} - 3 α_{g}] Δ θ

= 50 [180 \times 1 0^{- 6} - 3 \times 9 \times 1 0^{- 6}] (38 - 18)

= 0.153 cc

Q. A glass flask is filled up to a mark with 50cc of mercury at 18∘C. If the flask and contents are heated to 38∘C, how much mercury will be above the mark? ( α for glass is 9×10−6,∘C and coefficient of real expansion of mercury is 180×10−6/∘C )

0.85 c c

0.46 c c

0.153 c c

0.05 c c

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by ΔV=V0​[γL​−γg​]Δθ=V[γL​−3αg​]Δθ =50[180×10−6−3×9×10−6](38−18) =0.153cc

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by
$Δ V = V_{0} [γ_{L} - γ_{g}] Δ θ = V [γ_{L} - 3 α_{g}] Δ θ$
$= 50 [180 \times 1 0^{- 6} - 3 \times 9 \times 1 0^{- 6}] (38 - 18)$
$= 0.153 cc$