Question

Atmospheric pressure is equal to the weight of a vertical column of air, extending all the way up through the atmosphere, divided by the cross- sectional area of the column. (a) Explain why that must be true. [Hint: Apply Newton's second law to the column of air.] (b) If the air all the way up had a uniform density of \(1.29 \mathrm{kg} / \mathrm{m}^{3}\) (the density at sea level at \(0^{\circ} \mathrm{C}\) ), how high would the column of air be? (c) In reality, the density of air decreases with increasing altitude. Does that mean that the height found in (b) is a lower limit or an upper limit on the height of the atmosphere?

Short answer

(a) Atmospheric pressure equals the weight of the air column divided by the area due to Newton's second law. (b) Height is 7990 m. (c) It is an upper limit since real atmospheric density decreases with altitude.

Step-by-step solution

Understanding Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of air above a given point. Newton's second law (F = ma) shows that the force exerted by the air column must balance the gravitational force on the column of air. Thus, atmospheric pressure \( P \) is equal to the weight of the air column (force due to gravity) divided by the cross-sectional area \( A \).

Using the Concept of Density

If the air had a uniform density \( \rho = 1.29 \, \text{kg/m}^3 \), the weight \( W \) of the column would be \( W = \rho V g = \rho Ahg \), where \( V \) is the volume, \( h \) is the height, and \( g \) is the gravitational acceleration (approximately \(9.81 \, \text{m/s}^2\)). The pressure \( P \) is given by \( P = \frac{W}{A} = \rho hg \).

Calculating the Height of the Air Column

We know that at sea level, atmospheric pressure \( P \) is approximately \(101325 \, \text{Pa}\). Thus, using the equation \( P = \rho hg \), we can solve for \( h \): \( h = \frac{P}{\rho g} = \frac{101325}{1.29 \times 9.81} \). Calculating this gives \( h \approx 7990 \, \text{m}\).

Understanding the Effect of Decreasing Density with Altitude

In reality, air density decreases with altitude, resulting in a shorter column of air contributing the same pressure at sea level. This implies that the height calculated in (b), assuming uniform density, represents an upper limit because the decreasing density means a shorter column of air weighs the same as a taller column of constant density air.

Key concepts

Density

Density is a key concept when talking about atmospheric pressure. It refers to the mass of a substance per unit volume, usually expressed in kilograms per cubic meter (kg/m³). In the context of atmospheric pressure, density (\(\rho\)) tells us how much air is packed into a given space. For instance, at sea level and at 0°C, air density is about 1.29 kg/m³.

Knowing the density allows us to calculate other important properties, like pressure. If we assume a uniform density throughout a column of air, we can easily compute the weight of that column using the formula:

• \[ W = \rho V g = \rho Ahg \]

where \( W \) is the weight, \( \rho \) is the air density, \( V \) is the volume of the air column, \( A \) is the cross-sectional area of the column, \( h \) is the height of the air column, and \( g \) is the acceleration due to gravity (9.81 m/s²).

Understanding the impact of density on weight and pressure is crucial to understanding atmospheric pressure.

Newton's Second Law

Newton's Second Law is essential to understanding why atmospheric pressure is what it is. This law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. It is commonly written as:

• \[ F = ma \]

In terms of atmospheric pressure, the column of air above us is subject to gravity pulling it towards the Earth.

Applying Newton's Second Law, the force of gravity acting on this air column (its weight) is balanced by the atmospheric pressure exerted by the air on the surface below it. Essentially, the pressure is the result of the column's weight spread over its cross-sectional area, making the relationship:

• \[ P = \frac{W}{A} \]

where \( P \) is the pressure, \( W \) is the weight of the air column, and \( A \) is the cross-sectional area. This illustrates how Newton's Second Law helps in explaining the phenomena of atmospheric pressure.

Gravitational Force

Gravitational force plays a vital role in atmospheric pressure. It's the force that pulls the air column down towards Earth. This force is a significant reason why air at sea level is denser than at higher altitudes.

The gravitational force (\(F_g\)) on the column of air can be expressed as:

• \[ F_g = mg \]

Here, \( m \) represents the mass of the air column, and \( g \) is the gravitational acceleration (approximately 9.81 m/s²). This force creates the air column's weight, directly influencing atmospheric pressure. As height increases, gravitational pull decreases, which means the air becomes less dense.

This understanding helps explain why atmospheric pressure decreases with altitude – the weight of the air column reduces due to a reduction in gravitational force.

Sea Level Pressure

Sea level pressure is a standard reference point for atmospheric pressure and is crucial in understanding how pressure changes with altitude. At sea level, atmospheric pressure is approximately 101,325 Pascals (Pa). This pressure results from the weight of the air above pressing down on the surface.

When calculating atmospheric phenomena, assuming sea level pressure gives a baseline from which variations due to altitude changes can be understood. For instance, using the uniform air density and sea level pressure, the height of an air column can be calculated:

• \[ h = \frac{P}{\rho g} \]

In these formulas, symbols take on familiar meanings, allowing us to predict and comprehend pressure conditions at various altitudes. Adjusting calculations for real-world decreasing density scenarios help inform accurate atmospheric models. Understanding sea level pressure sets a foundation for more complex atmospheric studies.