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

Approximately
`3.81\; \rm m`.

Step-by-step explanation:

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

• Density of carbon tetrachloride: approximately
  `1.59* 10^(3)\; \rm kg \cdot m^(-3)`.
• Density of glycerol: approximately
  `1.26* 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* 10^(3) \; \rm kg \cdot m^(-3) * 9.81\; \rm N \cdot kg^(-1) * 3.02 \; \rm m \approx 4.71 * 10^(4) \; \rm N \cdot m^(-2) \end{aligned}`.

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

`\displaystyle h = (P)/(\rho \cdot g)`.

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

`\begin{aligned}h &= (P)/(\rho \cdot g) \\ &\approx (4.71 * 10^(4)\; \rm N \cdot m^(-2))/(1.26 * 10^(3)\; \rm kg \cdot m^(-3) * 9.81\; \rm N \cdot kg^(-1)) \approx 3.81\; \rm m\end{aligned}`.