A long capillary tube of uniform bore contains a thread of mercury which at 0ºc is 100 cm long and at 100ºc it is 101.55 cm. If the coefficient of real expansion of mercury is 1.82 × 10–4 ºc–1, find the coefficient of linear expansion of glass.
Answers
The coefficient of linear expansion of glass is 1.35×10⁻⁵ °C⁻¹.
Let V be the volume of mercury at 0°C,
and, V' be the volume of mercury at 100°C.
m = difference between the 2 temperatures = (100-0)°C
Let A be the cross sectional area of the tube at 0°C and
A' be the cross sectional area of the tube at 100°C.
Let the length of the mercury thread be l' at 100°C.
l' = V'/A' = [V(1+γm)]/[A(1+βm)]
where γ is the real coefficient of expansion
As V = Al
So,
l' = l(1+γm)]×[(1+βm)]⁻¹
= l(1+γm)]×[(1-βm)]
= l[1+(γ-β)m] [Taking terms common and neglecting smaller terms]
= l[1+(γ - 2δ/3)m]
where δ is the cubical expansion of glass
As, cubical expansion is 3 times that of linear expansion
So, δ = 3α, where α is the linear expansion of glass.
So, l' = l[1+(γ - 2α)m]
Given, l' = 101.55cm = 1.0155m [1cm = 0.01m]
l = 100cm = 1m
γ = 1.82 × 10⁻⁴ °C⁻¹
m = 100
Putting their respective values, we get:
1.0155 = [1 + (1.82 × 10⁻⁴ - 2α)100]
⇒1.0155 = [1 + 1.82×10⁻² - 200α]
⇒200α = 0.0027
⇒α = 0.0027/200 = 1.35×10⁻⁵ °C⁻¹
This is the coefficient of linear expansion.