Question

Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?

Answer 1

Answer:

Approximately $3.81\; \rm m$.

Explanation:

Look up the density $\rho$ of carbon tetrachloride, $\rm CCl_4$, and glycerol:

• Density of carbon tetrachloride: approximately $1.59\times 10^{3}\; \rm kg \cdot m^{-3}$.
• Density of glycerol: approximately $1.26\times 10^{3}\; \rm kg \cdot m^{-3}$.

Let $g$ denote the gravitational field strength. (Typically $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) For a column of liquid with a height of $h$, if the density of the liquid is $\rho$, the pressure at the bottom of the column would be:

$P = \rho\cdot g \cdot h$.

The pressure at the bottom of this carbon tetrachloride column would be:

$\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}$.

Rearrange the equation $P = \rho\cdot g \cdot h$ for $h$:

$\displaystyle h = \frac{P}{\rho \cdot g}$.

Apply this equation to calculate the height of the liquid glycerol column:

$\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}$.