Question

Determine the atmospheric pressure at a location where the barometric reading is \(750 \mathrm{mmHg}\). Take the density of mercury to be \(13,600 \mathrm{kg} / \mathrm{m}^{3}\)

Short answer

Answer: The atmospheric pressure at this location is approximately 100,361 Pa.

Step-by-step solution

Convert mm Hg to meters mercury (m Hg)

First, we need to convert the given barometric reading from millimeters of Mercury (mm Hg) to meters of Mercury (m Hg). In this case, the barometric reading is \(750 \mathrm{mmHg}\). To convert this to meters, simply divide by 1000:$$750 \mathrm{mmHg} * \frac{1 \mathrm{mHg}}{1000 \mathrm{mmHg}} = 0.75 \mathrm{mHg}$$So, the barometric reading in meters is \(0.75 \mathrm{mHg}\).

Use the formula for pressure to calculate atmospheric pressure in Pascals

Now that we have the height of the mercury column in meters, we can use the formula for pressure to calculate the atmospheric pressure in Pascals. The formula for pressure is given by:$$P = \rho gh$$Where \(P\) is the pressure, \(\rho\) is the density of the fluid (in this case, mercury), \(g\) is the gravitational constant, and \(h\) is the height of the fluid column. We know the density of mercury is \(13,600 \mathrm{kg} / \mathrm{m}^{3}\), the gravitational constant is \(9.81 \mathrm{m} / \mathrm{s}^{2}\), and the height of the mercury column is \(0.75 \mathrm{mHg}\). Plugging these values into the formula, we get:$$P = 13600 \mathrm{kg} / \mathrm{m}^{3} * 9.81 \mathrm{m} / \mathrm{s}^{2} * 0.75 \mathrm{mHg}$$

Calculate the final value for atmospheric pressure in Pascals

Now, we can simply perform the multiplication to find the atmospheric pressure in Pascals:$$P = 13600 * 9.81 * 0.75 = 100361 \mathrm{Pa}$$So, the atmospheric pressure at the location with a barometric reading of \(750 \mathrm{mmHg}\) is approximately \(100,361 \mathrm{Pa}\).

Key concepts

Barometric Reading Conversion

Understanding how to convert barometric readings is essential in exercises dealing with atmospheric pressure. The barometer is an instrument that measures atmospheric pressure, often provided in millimeters of mercury (mmHg). However, scientific calculations typically require these measurements in meters (mHg) since the standard unit for measuring pressure is Pascals (Pa). The conversion is straightforward: divide the mmHg value by 1000 to get the measurement in meters of mercury (mHg). For example, a reading of 750 mmHg is equivalent to 0.75 mHg after conversion.

It's crucial for students to remember the conversion factor and use it correctly to ensure accurate calculations in physics and related fields. This step might seem minor, but it is the foundation for obtaining the right values when applying the pressure formula later on. Simplifying units early on can prevent confusion and mistakes in subsequent steps of the problem-solving process.

Density of Mercury

Mercury is a unique element with a high density, which is why it has been traditionally used in barometers and thermometers. Its density is significant when calculating pressure using the formula \( P = \rho gh \), where \( \rho \) represents the density of the fluid. For mercury, this value is often provided as 13,600 kg/m³.

The density of mercury, combined with its relative incompressibility, allows for the precise measurement of atmospheric pressure. When dealing with related exercises, understanding that the density value is a constant provided in the problem—and ensuring that it is in the correct units of kilograms per cubic meter (kg/m³)—is imperative. Always verify that the density is consistent with the units used in the pressure formula to avoid discrepancies in the final result.

Pressure Formula Application

The atmospheric pressure calculation is based on the fundamental pressure formula \( P = \rho gh \). In this equation, \( P \) represents the pressure exerted by the fluid, \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid column. It is essential to apply this formula correctly, using the units of \( \rho \) (kg/m³), \( g \) (m/s²), and \( h \) (m).

The student is advised to methodically match each term with its provided or known value and perform the multiplication. The product will provide the pressure in Pascals (Pa), the SI unit for pressure. Remembering to align units across all elements in the formula is key, and doing this step methodically aids in internalizing the concept for application in varied contexts.

Gravitational Constant

The constant \( g \) in the pressure formula denotes the acceleration due to gravity, universally accepted as 9.81 m/s² on Earth's surface. This value is crucial for accurately determining the force exerted by gravity on a fluid, which in turn affects pressure calculations. Gravity's consistency at the Earth's surface allows this value to be treated as a constant in these types of problems.

When students deal with the gravitational constant in physics problems, it's important to highlight its role and standard value. While \( g \) remains relatively unchanged within the context of Earth-bound problems, it's also interesting to note that it would differ on other celestial bodies, an aspect that can be explored in more advanced physics education. For most high school or introductory level physics problems, this sophisticated concept can serve as a stepping stone towards understanding how gravity impacts different physical phenomena.