Question

Normal atmospheric pressure is $\mathbf{1}\mathbf{.}\mathbf{013}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{\text{Pa}}$. The approach of a storm causes the height of a mercury barometer to drop by$\mathbf{\text{20.0mm}}$from the normal height. What is the atmospheric pressure?

Short answer

The pressure interred is$P=98.6\text{\hspace{0.17em}kPa}$

Step-by-step solution

Identification of the given data:

The given data can be listed below as

• Atmospheric pressure,$P_0=1.013\times 10{}^5\text{\hspace{0.17em}Pa}$
• Density of mercury, $\rho =13600\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$
• The height,role="math" localid="1663736570947" $h=-20\times {10}^{-3}\text{m}$

Pressure Measurement:

Pascal’s law states that, when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.

The pressure in a fluid at rest varies with depthin the fluid according to the expression,

$\mathit{P}\mathbf{=}{\mathit{P}}_{\mathbf{0}}\mathbf{+}\mathit{\rho }\mathit{g}\mathit{h}$

Here, ${\mathit{P}}_{\mathbf{0}}$is the pressure at $\mathit{h}\mathbf{=}\mathbf{0}$, $\mathit{g}$is the acceleration due to gravity having a value $\mathbf{\text{10}}\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$}\right.$,and $\mathit{\rho }$is the density of the fluid.

Evaluating atmospheric pressure:

Now use Pascal’s equation to find the pressure that causes a drop$h$in the mercury barometer.

$P=P_0+\rho gh$

Substituting the all given value in above equation,

$\begin{array}{rcl}P& =& 1.013\times {10}^5\text{\hspace{0.17em}Pa}+13600\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\times \text{10}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\times -20\times {10}^{-3}\text{\hspace{0.17em}m}\\ & =& 1.013\times {10}^5\text{\hspace{0.17em}Pa}+[-272\times {10}^4\times {10}^{-3}]\text{Pa}\\ & =& 9.86\times {10}^4\text{Pa}\\ & =& 98.6\text{\hspace{0.17em}kPa}\end{array}$

Hence, the atmospheric pressure is$98.6\text{\hspace{0.17em}kPa}$.