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Question: A specific gravity bottle is completely filled by 271.92 g of mercury at \({0^ \circ }C\) the mass o...

A specific gravity bottle is completely filled by 271.92 g of mercury at 0∘C{0^ \circ }C the mass of mercury which would fill in specific gravity bottle at 100∘C{100^ \circ }C is (Coefficient of linear expansion of the material of the specific gravity bottle 8.0×10−6/∘C8.0 \times {10^{ - 6}}{/^ \circ }C and coefficient of areal expansion of mercury = 0.000182/∘C0.000182{/^ \circ }C)
A. 267 g
B. 277 g
C. 500 g
D. 300 g

Explanation

Solution

The density is defined as the amount of mass per unit volume of the substance –
ρ=MV\rho = \dfrac{M}{V}
The specific gravity bottle is used to measure the specific gravity of a liquid, which is the measure of the number of times its density is higher than that of water.

Complete step by step solution:
Step 1: List the given data –
Mass of mercury at 0∘C{0^ \circ }C, m1{m_1}= 271.92 g
Coefficient of linear expansion of the material of specific gravity bottle, αg=8.0×10−6/∘C{\alpha _g} = 8.0 \times {10^{ - 6}}{/^ \circ }C
Coefficient of areal expansion of mercury, β=0.000182/∘C\beta = 0.000182{/^ \circ }C
To find –
Mass of mercury after heating it to 100∘C,m2=?{100^ \circ }C,{m_2} = ?
The mass of mercury at 100∘C,m2=ρ2V2{100^ \circ }C,{m_2} = {\rho _2}{V_2}
where ρ2&V2{\rho _2}\& {V_2} are density and volume of the mercury at the temperature respectively.
Thus, finding the value of the density and volume at 100∘C{100^ \circ }Cwill give us the mass.
Step 2: Find the value of ρ2{\rho _2}
The mercury expands on heating from 0∘C{0^ \circ }Cto 100∘C{100^ \circ }C. The new density after expansion is given by,
ρ2=ρ1(1−β(Δθ)){\rho _2} = {\rho _1}\left( {1 - \beta \left( {\Delta \theta } \right)} \right)
where Δθ&β\Delta \theta \& \beta are the increase in temperature and coefficient of areal expansion, respectively.
Also,
ρ1=m1V1 Substituting, ρ1=271.92V1  {\rho _1} = \dfrac{{{m_1}}}{{{V_1}}} \\\ Substituting, \\\ {\rho _1} = \dfrac{{271.92}}{{{V_1}}} \\\
Substituting the values of ρ1{\rho _1}, Δθ&β\Delta \theta \& \beta in the equation above;

ρ2=ρ1(1−β(Δθ)) ρ2=271.92V1(1−(0.000182)(100−0)) Solving ρ2=271.92V1(1−0.000182×100) ρ2=271.92V1(1−0.0182) ρ2=271.92×0.9818V1 ρ2=266.97V1−(1)  {\rho _2} = {\rho _1}\left( {1 - \beta \left( {\Delta \theta } \right)} \right) \\\ {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - \left( {0.000182} \right)\left( {100 - 0} \right)} \right) \\\ Solving \\\ {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - 0.000182 \times 100} \right) \\\ {\rho _2} = \dfrac{{271.92}}{{{V_1}}}\left( {1 - 0.0182} \right) \\\ {\rho _2} = \dfrac{{271.92 \times 0.9818}}{{{V_1}}} \\\ {\rho _2} = \dfrac{{266.97}}{{{V_1}}} - \left( 1 \right) \\\

Step 3: Find the value of V2{V_2}
There is a volumetric expansion on heating the specific gravity bottle from 0∘C{0^ \circ }Cto 100∘C{100^ \circ }C. The new density after expansion is given by,
V2=V1(1+γg(Δθ)){V_2} = {V_1}\left( {1 + {\gamma _g}\left( {\Delta \theta } \right)} \right)
where Δθ&γg\Delta \theta \& {\gamma _g} are the increase in temperature and coefficient of volumetric expansion, respectively.
Coefficient of volumetric expansion, γg=3×αg{\gamma _g} = 3 \times {\alpha _g}
Hence, the equation becomes –
V2=V1(1+3αg(Δθ)){V_2} = {V_1}\left( {1 + 3{\alpha _g}\left( {\Delta \theta } \right)} \right)
Substituting the values, we get –
V2=V1(1+3αg(Δθ)) V2=V1(1+3×8×10−6(100−0)) Solving, V2=V1(1+3×8×10−6×100) V2=V1(1+3×8×10−6×102) V2=V1(1+3×8×10−4) V2=V1(1+24×10−4) V2=V1(1+0.0024) ∴V2=V1(1.0024)−(2)  {V_2} = {V_1}\left( {1 + 3{\alpha _g}\left( {\Delta \theta } \right)} \right) \\\ {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}}\left( {100 - 0} \right)} \right) \\\ Solving, \\\ {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}} \times 100} \right) \\\ {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 6}} \times {{10}^2}} \right) \\\ {V_2} = {V_1}\left( {1 + 3 \times 8 \times {{10}^{ - 4}}} \right) \\\ {V_2} = {V_1}\left( {1 + 24 \times {{10}^{ - 4}}} \right) \\\ {V_2} = {V_1}\left( {1 + 0.0024} \right) \\\ \therefore {V_2} = {V_1}\left( {1.0024} \right) - \left( 2 \right) \\\
Step 4: Combining (1)\left( 1 \right) and (2)\left( 2 \right), we get the mass of the mercury –
m2=ρ2V2 Substituting, m2=(266.97V1)(V1(1.0024)) m2=266.97×1.0024 m2=267.61g  {m_2} = {\rho _2}{V_2} \\\ Substituting, \\\ {m_2} = \left( {\dfrac{{266.97}}{{{{{V}}_1}}}} \right)\left( {{{{V_1}}}\left( {1.0024} \right)} \right) \\\ {m_2} = 266.97 \times 1.0024 \\\ {m_2} = 267.61g \\\
Therefore, the correct option is OPTION C.

Note:
There are three types of coefficients of expansion, namely, Linear (α)\left( \alpha \right), Superficial or Areal (β)\left( \beta \right) and Volumetric (γ)\left( \gamma \right). The relationship between them is as follows:
α1=β2=γ3\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}