Question

If on heating liquid through $80^\circ C$ , the mass expelled is (1/100)$^{th}$ of mass still remaining, the coefficient of apparent expansion of liquid is
1)$1.25 \times {10^{ - 4}}$/$^\circ C$
2) $12.5 \times {10^{ - 4}}$/$^\circ C$
3) $1.25 \times {10^{ - 5}}$/$^\circ C$
4) None of these

Answer 1

Hint: Coefficient of the apparent expansion is represented as ${\gamma _{app}}$ and by using the equation of the expansion we have relation between mass expelled, mass remaining, coefficient of expansion, and temperature change as well. So using it we can find the desired answer.

Complete Step by Step Solution
Correct answer: $1.25 \times {10^{ - 4}}^\circ C$
Coefficient of apparent expansion : It is defined as the ratio of the apparent change in volume of the liquid to its original volume per $1^\circ C$ rise. It’s S.I. unit is $^\circ {C^{ - 1}}$ or ${K^{ - 1}}$.
Coefficient of real expansion: It is defined as the ratio of real change in volume to its original volume per $1^\circ C$ rise in temperature. It’s S.I. unit is $^\circ {C^{ - 1}}$ or ${K^{ - 1}}$.
On heating the atoms do not expand but the volume they take up does.
As we want to find coefficient of apparent expansion so,
Formula used would be
${\gamma _{app}} = \dfrac{{{m_e}}}{{{m_r} \times \Delta T}}$
Remember this formula is derived from $\Delta V = {V_o}({\gamma _L} - {\gamma _C})\Delta T = {V_o}({\gamma _{app}})\Delta T$
Where ${\gamma _{app}} = {\gamma _L} - {\gamma _C}$
And
${\gamma _L}$ is the cubical constant of liquid.
${\gamma _C}$ is the cubical constant of a container.
Where ${m_e}$ is mass expelled, ${m_r}$ is mass remaining and $\Delta T$ is change in temperature.
So
As given
$\Delta T = 80$
And
mass expelled$ = \dfrac{1}{{100}}mass$remaining
${m_e} = \dfrac{1}{{100}}{m_r}$
Hence,
$\dfrac{{{m_e}}}{{{m_r}}} = \dfrac{1}{{100}}$
Now using both values we get
${\gamma _{app}} = \dfrac{1}{{100 \times 80}} = 1.25 \times {10^{ - 4}}$
So, Coefficient of the apparent expansion ${\gamma _{app}} = 1.25 \times {10^{ - 4}}^\circ C$

Note
 We can also solve it by using formula $\Delta V = {V_o}({\gamma _{app}})\Delta T$
Where $\Delta T$ would be change in temperature
${V_o}$ would be the total mass ( ${m_e} + {m_r}$ ).
$\Delta V$ would be mass expelled (${m_r}$ )
Where total mass can be also taken as any arbitrary variable (x).