A glass flask whose volume is 1000.00 cm^3 at 0.0∘C is completely filled with mercury at this temperature. When flask and mercury are warmed to 55.1 ∘C, 8.71 cm^3 of mercury overflow
Compute the coefficient of volume expansion of the glass. (The coefficient of volume expansion of the mercury is 18×10^−5 K^−1.)
$Given-\phantom{\rule{0ex}{0ex}}{V}_{0}=1000c{m}^{3}\phantom{\rule{0ex}{0ex}}{T}_{1}={0}^{0}C\phantom{\rule{0ex}{0ex}}{T}_{2}=55.{1}^{0}C\phantom{\rule{0ex}{0ex}}Thecoefficientofvolumeexpansionofthemercury{\beta}_{Hg}=18\times {10}^{-5}{K}^{-1}\phantom{\rule{0ex}{0ex}}OverflowofHg=8.71c{m}^{3}$
$Solution-\phantom{\rule{0ex}{0ex}}\delta {V}_{Hg}-\delta {V}_{flack}=8.71c{m}^{3}---------\left(1\right)\phantom{\rule{0ex}{0ex}}Wehave\phantom{\rule{0ex}{0ex}}\delta {V}_{Hg}={V}_{0}\times {\beta}_{Hg}\times \delta T\phantom{\rule{0ex}{0ex}}=1000\times 18\times {10}^{-5}\times {(55.1-0)}^{0}C=9.918c{m}^{3}\phantom{\rule{0ex}{0ex}}\delta {V}_{flask}=\delta {V}_{Hg}-8.71=9.918-8.71=1.208c{m}^{3}\phantom{\rule{0ex}{0ex}}Finallytocalculate{\beta}_{flask}\phantom{\rule{0ex}{0ex}}Wehave\phantom{\rule{0ex}{0ex}}{\beta}_{flask}=\frac{\delta {V}_{flask}}{{V}_{0}\times \delta T}\phantom{\rule{0ex}{0ex}}=\frac{1.208}{1000\times {(55.1-0)}^{0}C}\phantom{\rule{0ex}{0ex}}=2.192\times {10}^{-5}{\left({C}^{0}\right)}^{-1}AswecanseetheflowofHghappenedasaresultofagreatvolumethermalcoefficient\beta thantheflask.$