The densities of a liquid at 0^{circ} C and 1 | vedclass

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(B) Density at $0^{\circ} C, \rho_0 = 1.0127 \ g/cm^3$Density at $100^{\circ} C, \rho_{100} = 1 \ g/cm^3$Coefficient of real expansion of liquid, $\ga

Vedclass · Accepted Answer

$\frac{3}{1.01}$

Vedclass · Answer

(B) Density at $0^{\circ} C, ho_0 = 1.0127 \ g/cm^3$Density at $100^{\circ} C, ho_{100} = 1 \ g/cm^3$Coefficient of real expansion of liquid, $\gamma_{	ext{real}} = \frac{ho_0 - ho_{100}}{ho_{100} 	imes \Delta t}$$\gamma_{	ext{real}} = \frac{1.0127 - 1}{1 	imes 100} = \frac{0.0127}{100} = 1.27 	imes 10^{-4} /^{\circ} C$Coefficient of volume expansion of glass, $\gamma_g = 3 \alpha = 3 	imes 9 	imes 10^{-6} = 27 	imes 10^{-6} = 0.27 	imes 10^{-4} /^{\circ} C$Coefficient of apparent expansion, $\gamma_{	ext{app}} = \gamma_{	ext{real}} - \gamma_g$$\gamma_{	ext{app}} = 1.27 	imes 10^{-4} - 0.27 	imes 10^{-4} = 1 	imes 10^{-4} /^{\circ} C$Using the formula for apparent expansion: $\gamma_{	ext{app}} = \frac{\Delta m}{m_2 \Delta t}$, where $\Delta m$ is the mass expelled and $m_2$ is the remaining mass.Alternatively, $\Delta m = m_1 	imes \gamma_{	ext{app}} 	imes \Delta t$ (approximate for small expansion) or more accurately:$V_0 = \frac{m_1}{ho_0}$. At $100^{\circ} C$, $V_{100} = V_0(1 + \gamma_g \Delta t) = \frac{m_1}{ho_0}(1 + \gamma_g \Delta t)$.Mass remaining $m_2 = V_{100} 	imes ho_{100} = \frac{m_1 ho_{100}}{ho_0}(1 + \gamma_g \Delta t) = \frac{300 	imes 1}{1.0127}(1 + 27 	imes 10^{-6} 	imes 100) \approx 296.24 \ g$.Mass expelled $= m_1 - m_2 = 300 - \frac{300}{1.01} = \frac{3}{1.01} \ g$.

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