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(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.$\Delta V = V_0 [\gamma_m -

Vedclass · Accepted Answer

$15.2$

Vedclass · Answer

(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.$\Delta V = V_0 [\gamma_m - \gamma_g] \Delta 	heta$Since $\gamma_g = 3\alpha_g$,we have:$\Delta V = V_0 [\gamma_m - 3\alpha_g] \Delta 	heta$Given:$V_0 = 1 \ L = 1000 \ cc$$\gamma_m = 1.82 	imes 10^{-4} /{ }^{\circ} C = 182 	imes 10^{-6} /{ }^{\circ} C$$\alpha_g = 0.1 	imes 10^{-4} /{ }^{\circ} C = 10 	imes 10^{-6} /{ }^{\circ} C$$\Delta 	heta = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$Substituting the values:$\Delta V = 1000 	imes [182 	imes 10^{-6} - 3(10 	imes 10^{-6})] 	imes 100$$\Delta V = 1000 	imes [182 	imes 10^{-6} - 30 	imes 10^{-6}] 	imes 100$$\Delta V = 1000 	imes [152 	imes 10^{-6}] 	imes 100 = 15.2 \ cc$

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