Question

The coefficient of volume expansion of a certain
olive oil is 0.68 x 10-3 K-1. A 1.0-L glass beaker is
filled to the brim with olive oil at room temperature.
The beaker is placed on a range and the temperature
of the oil and beaker increases by 25°C. As a result,
0.0167 L of olive oil spills over the top of the
beaker. Which of the following values is closest to
the coefficient of linear expansion of the glass from
which the beaker is made?
(a) 1 x 10-6 K-1
(c) 1 x 10-5 K-1
(b) 4 x 10-6 K-1
(d) 2 x 10-5 K-1​

Answer 1

Given:

βₙ =  0.68 x 10-3

T = 25°C

ΔV₀ - ΔVₙ = 0.0167 L

To find:

coefficient of linear expansion of the glass from  which the beaker is made

Solution:

co efficient of volume expansion of gass

ΔVₙ = V₀  βₙ ΔT

co efficient of volume expansion of oil

ΔV₀ = V₀ β₀ ΔT

Subtracting both the equations we get

ΔV₀ - ΔVₙ = ΔV₀  βₙ ΔT - V₀ β₀ ΔT

ΔV₀ - ΔVₙ = 0.0167

V₀ = 1.06 L

Substituting we get,

(0.0167)βₙ ΔT - (0.0167) β₀ ΔT = 0.0167

βₙ = $\frac{\beta _0\triangle T - 0.0167}{\triangle T}$

β₀ = 0.68 × 10⁻³

ΔT = 25⁰ C

βₙ = $\frac{0.68 . 10^-^3 . (273+25) -0.0167}{273+25}$

βₙ = 6.24 ×10⁻⁴K⁻¹

Therefore,

βₙ = 3αₙ

αₙ = $\frac{\beta _n}{3}$

αₙ = $\frac{6.24 * 10^-^4}{3}$

αₙ = 2.08 × 10⁻⁴ K⁻¹