Question

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic expansivity of mercury as equal to 180 × 10-6 K -1

Answer 1

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$