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Question: If on heating liquid through \(80^\circ C\) , the mass expelled is (1/100)\(^{th}\) of mass still re...

If on heating liquid through 80∘C80^\circ C , the mass expelled is (1/100)th^{th} of mass still remaining, the coefficient of apparent expansion of liquid is
1)1.25×10−41.25 \times {10^{ - 4}}/∘C^\circ C
2) 12.5×10−412.5 \times {10^{ - 4}}/∘C^\circ C
3) 1.25×10−51.25 \times {10^{ - 5}}/∘C^\circ C
4) None of these

Explanation

Solution

Coefficient of the apparent expansion is represented as γapp{\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: 1.25×10−4∘C1.25 \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∘C1^\circ C rise. It’s S.I. unit is ∘C−1^\circ {C^{ - 1}} or K−1{K^{ - 1}}.
Coefficient of real expansion: It is defined as the ratio of real change in volume to its original volume per 1∘C1^\circ C rise in temperature. It’s S.I. unit is ∘C−1^\circ {C^{ - 1}} or K−1{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
γapp=memr×ΔT{\gamma _{app}} = \dfrac{{{m_e}}}{{{m_r} \times \Delta T}}
Remember this formula is derived from ΔV=Vo(γL−γC)ΔT=Vo(γapp)ΔT\Delta V = {V_o}({\gamma _L} - {\gamma _C})\Delta T = {V_o}({\gamma _{app}})\Delta T
Where γapp=γL−γC{\gamma _{app}} = {\gamma _L} - {\gamma _C}
And
γL{\gamma _L} is the cubical constant of liquid.
γC{\gamma _C} is the cubical constant of a container.
Where me{m_e} is mass expelled, mr{m_r} is mass remaining and ΔT\Delta T is change in temperature.
So
As given
ΔT=80\Delta T = 80
And
mass expelled=1100mass = \dfrac{1}{{100}}massremaining
me=1100mr{m_e} = \dfrac{1}{{100}}{m_r}
Hence,
memr=1100\dfrac{{{m_e}}}{{{m_r}}} = \dfrac{1}{{100}}
Now using both values we get
γapp=1100×80=1.25×10−4{\gamma _{app}} = \dfrac{1}{{100 \times 80}} = 1.25 \times {10^{ - 4}}
So, Coefficient of the apparent expansion γapp=1.25×10−4∘C{\gamma _{app}} = 1.25 \times {10^{ - 4}}^\circ C

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