The pressure unit torr is defined as the pressure that will support a column of mercury \(1 \mathrm{mm}\) high. Meteorologists often give barometric pressure in inches of mercury, defined analogously. Express each of these in SI units. (Hint: Mercury's density is \(\left.1.36 \times 10^{4} \mathrm{kg} / \mathrm{m}^{3} .\right)\)

Understanding what these measurements really mean forms the basis for being able to convert them to an SI unit. Both torr and inches of mercury are measurements of pressure based on the height of a column of mercury. 1 torr is the amount of pressure necessary to raise a column of mercury 1 millimeter, and similarly for inches of mercury.

Pressure is calculated using the formula \( P = \frac{F}{A} \), where P is pressure, F is force and A is area. Since force is equal to mass times acceleration, \( F = m \cdot g \), and mass is equal to density times volume, \( m = \rho \cdot V \), we can substitute these equations into the pressure equation to give \( P_{torr} = \rho_{Hg} \cdot g \cdot h_{torr} \) in terms of SI units, where \( \rho_{Hg} \) is the density of Mercury (\( 1.36 \times 10^{4} \mathrm{kg/m^{3}} \)), g is the acceleration due to gravity (\(9.8 \mathrm{m/s^{2}}\)) and \( h_{torr} \) is the height of mercury expressed in meters (for 1 torr, \( h_{torr} = 1 \mathrm{mm} = 0.001 \mathrm{m} \)). Evaluating this expression will give us the value of 1 torr in Pascals (Pa).

To convert inches of mercury to Pascals, the same formula can be used as in Step 2, changing the height \( h_{\inch} \) to the corresponding value in meters (for 1 inch, \( h_{\inch} = 1 \mathrm{inch} \approx 0.0254 \mathrm{m} \)).

Applying the above approach for calculating pressure, the value for 1 torr in Pascals (Pa) comes out by substituting the corresponding values to be \( P_{torr} \approx 1.36 \times 10^{4} \mathrm{kg/m^{3}} \times 9.8 \mathrm{m/s^{2}} \times 0.001 \mathrm{m} \approx 133 \, Pa \). Similarly, the pressure for 1 inch of mercury comes out to be \( P_{\inch} = 1.36 \times 10^{4} \mathrm{kg/m^{3}} \times 9.8 \mathrm{m/s^{2}} \times 0.0254 \mathrm{m} \approx 3386 \, Pa \).