The density of a substance at 20∘C and 120∘C are 1.1g c.c

Question

The density of a substance at

20^{\circ} C

and

120^{\circ} C

are

1.1 g c . c^{- 1}

and

1 g c . c^{- 1}

respecttively. Find the coefficient of cubical expansion of the solid.

Options:

A . 0.001∘C−1

B . 0.002∘C−1

C . 0.003∘C−1

D . 0.004∘C−1

Answer: Option A
:
A
Given: Initial density,

d_{1} = 1.1 g c . c^{- 1}

Final density,

d_{2} = 1 g c . c^{- 1}

Initial tempereature,

t_{1} = 20^{\circ} C

Final temperature,

t_{2} = 120^{\circ} C

Volumetric expansion of solids can be expressed as,

V_{2} = V_{1} (1 + γ (t_{2} - t_{1}))

where,

V_{1}

is the initial volume,

V_{2}

is the final volume,

t_{1}

is the initial temperature,

t_{2}

is the final temperature and

γ

is the coefficient of cubical expansion of solid.
We know

v o l u m e = \frac{m a s s}{d e n s i t y}

Let

d_{1}

be the density at volume

V_{1}

and

d_{2}

be the density at volume

V_{2}

Since mass of the substance,

m

remains same, we have,
or,

\frac{m}{d_{2}} = \frac{m}{d_{1}} (1 + γ (t_{2} - t_{1}))

or,

γ = \frac{d_{1} - d_{2}}{d_{2} (t_{2} - t_{1})}

or,

γ = \frac{1.1 - 1}{1 (120 - 20)} = \frac{0.1}{100} = 0.001

∴ γ = {0.001}^{\circ} C^{- 1}

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