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Question: The density of mercury is 13.6 \(gc{m^{ - 3}}\)at \({0^0}{\text{C}}\) and its coefficients of cubica...

The density of mercury is 13.6 gcm−3gc{m^{ - 3}}at 00C{0^0}{\text{C}} and its coefficients of cubical expansion is 1.82×10−4 C−11.82 \times {10^{ - 4}}{\text{ }}{{\text{C}}^{ - 1}}. Calculate the density of mercury at 500C{50^0}{\text{C}}.

Explanation

Solution

Hint – Density is defined as mass per unit volume, if volume is taken as unity then the density is equal to mass, hence consider mass of mercury as 13.6 gm for volume of 1cm31c{m^3}.

Complete step-by-step answer:
It is given that the density ρ\rho of mercury is 13.6 g cm3 at 00C{0^0}{\text{C}} and its coefficient of cubical expansion is 1.82×10−4 C−11.82 \times {10^{ - 4}}{\text{ }}{{\text{C}}^{ - 1}}.
Let the mass of mercury (m) = 13.6 gm.
And the volume (V) = 1 cm3.
And also let, α\alpha = 1.82×10−4 C−11.82 \times {10^{ - 4}}{\text{ }}{{\text{C}}^{ - 1}}
Now when we heated the mercury its volume changes so the change in volume (dv) when it is heated up to 500C is
dv=v×α×dtdv = v \times \alpha \times dt where (dt is change in temperature).
⇒dv=1×1.82×10−4×(50−0)\Rightarrow dv = 1 \times 1.82 \times {10^{ - 4}} \times \left( {50 - 0} \right)
⇒dv=91×10−4cm3\Rightarrow dv = 91 \times {10^{ - 4}}c{m^3}
Therefore, new volume V1 = V + dv = 1.0091 cm3.
Therefore, new density ρ\rho = (m/V1) = (13.6/1.0091) = 13.48 g cm-3.
So this is the required answer.

Note – In this question the coefficient of cubical expansion is given this means the mercury must be heated and its volume will eventually alter too, that’s why the change in volume is being taken into consideration. It is advised to remember the direct formula for change in volume in terms of coefficient of cubical expansion.