/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The pressure unit torr is define... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 18

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 \(\left.1.36 \times 10^{4} \mathrm{kg} / \mathrm{m}^{3} .\right)\)

Short Answer

Expert verified
When calculated in SI units, \(1 \) torr is approximately \( 133.32 \) pascals, and \( 1 \) in Hg is approximately \( 3386.39 \) pascals.

Step by step solution

01

Understand Pressure Calculation

Recall the definition of pressure, which is the force per unit area. It can be represented as: \( P = \frac{F}{A} \). In this situation, the force is due to the weight of the mercury column, which is the product of mass and gravity: \( F = m \cdot g \). Mass can be expressed by the product of volume and density ( \( \rho \) ): \( m = V \cdot \rho \) . We can substitute these equations back into our pressure formula.
02

Analyze Column Properties

The volume of the mercury column is the product of the base area \( A \) and height \( h \) of the column: \( V = A \cdot h \) . Given the height is \(1 \mathrm{mm}\) or \(1 \mathrm{in}\), we can use these values for \( h \). Substituting the volume equation into the mass equation and then into the pressure equation we get: \( P = \frac{m \cdot g}{A} = \frac{V \cdot \rho \cdot g}{A} = h \cdot \rho \cdot g \) .
03

Calculate Pressure SI Units

Convert the dimensions to SI units. For 1 torr with column height in \( \mathrm{mm} \), use \( h = 1 \times 10^{-3} \), and for metrics using inches, use \( h = 1 \times 2.54 \times 10^{-2} \) . In terms of \( g \), gravity in \( \mathrm{m/s^2} \) , we use \( g = 9.81 \) . Given \( \rho \) as \( 1.36 \times 10^4 \) kg/m鲁 for the density of mercury, plug all these values into the pressure equation to get the final pressure values in Pascals.

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