Question

The variation of pressure within the biosphere affects not only living things but also systems such as aircraft and undersea exploration vehicles. (a) Plot the variation of atmospheric pressure, in atm, versus elevation \(z\) above sea level, in \(\mathrm{km}\), ranging from 0 to \(10 \mathrm{~km}\). Assume that the specific volume of the atmosphere, in \(\mathrm{m}^{3} / \mathrm{kg}\), varies with the local pressure \(p\), in \(\mathrm{kPa}\), according to \(v=72.435 / p\). (b) Plot the variation of pressure, in atm, versus depth \(z\) below sea level, in \(\mathrm{km}\), ranging from 0 to \(2 \mathrm{~km}\). Assume that the specific volume of seawater is constant, \(v=0.956 \times\) \(10^{-3} \mathrm{~m}^{3} / \mathrm{kg}\) In each case, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and the pressure at sea level is \(1 \mathrm{~atm}\)

Short answer

Variation of atmospheric pressure is plotted using \( p(z) = e^{-0.03416 z} \), whereas underwater pressure is plotted using \( p(z) = 101.325 + 10.264 \times 10^6 \times z \).

Step-by-step solution

Atmospheric Pressure Formula

The pressure in the atmosphere decreases exponentially with elevation. The barometric formula can be written as: \( p(z) = p_0 e^{-\frac{mgz}{RT}} \) where \( p_0 \) is the sea level pressure, \( m \) is the molar mass of air, \( g \) is the acceleration due to gravity, \( z \) is the elevation, \( R \) is the universal gas constant, and \( T \) is the temperature.

Given Data for Atmospheric Pressure

Substitute the following values into the formula: \( g = 9.81 \, \mathrm{m/s^2} \), \( R = 8.314 \times 10^3 \, \mathrm{J/(kmol \, K)} \), \( m = 28.964 \, \mathrm{kg/kmol} \) (average molar mass of air), assume \( T = 288.15 \, \mathrm{K} \) (standard temperature), and \( p_0 = 101.325 \, \mathrm{kPa} \). Apply conversion: 1 atm = 101.325 kPa.

Derive and Simplify the Equation

Express \( p(z) \) in terms of \( z \). \[ p(z) = 101.325 \, e^{-\frac{9.81 \, 28.964 \, z}{8.314 \, 288.15}} \] Simplify the exponent: \( -\frac{9.81 \, 28.964}{8.314 \, 288.15} \approx -0.03416 \). The equation becomes: \[ p(z) = 101.325 \, e^{-0.03416 \, z} \]

Convert \( p(z) \) to atm

Divide by 101.325 to convert from kPa to atm: \[ p(z) = e^{-0.03416 \, z} \]

Plot the Variation of Atmospheric Pressure

Generate values for \( z = 0 \) to \( 10 \, \mathrm{km} \). Calculate \( p(z) \) for each \( z \) using the simplified formula and plot \( p(z) \) versus \( z \) (in atm).

Underwater Pressure Formula

The pressure underwater increases linearly with depth. The formula is: \( p(z) = p_0 + \rho g z \) where \( \rho \) is the density of seawater, \( g \) is the acceleration due to gravity, and \( z \) is the depth.

Given Data for Underwater Pressure

Substitute the following values into the formula: \( \rho = \frac{1}{0.956 \times 10^{-3}} \, \mathrm{kg/m^3} = 1046.45 \, \mathrm{kg/m^3} \), \( g = 9.81 \, \mathrm{m/s^2} \), and \( p_0 = 101.325 \, \mathrm{kPa} \).

Calculate the Pressure at Different Depths

Express \( p(z) \) in terms of \( z \). \[ p(z) = 101.325 + 1046.45 \times 9.81 \times 1000 \times z \] Simplify: \[ p(z) = 101.325 + 10.264 \times 10^6 \times z \] Convert kPa to atm for each calculation point.

Plot the Variation of Underwater Pressure

Generate values for \( z = 0 \) to \( 2 \, \mathrm{km} \). Calculate \( p(z) \) for each \( z \) using the underwater pressure formula and plot \( p(z) \) versus \( z \) (in atm).

Key concepts

atmospheric pressure

Atmospheric pressure is the force exerted by the weight of the air in the Earth's atmosphere. At sea level, this pressure is approximately 1 atm (101.325 kPa). As the elevation increases, the atmospheric pressure decreases because there's less air above pushing down. This can be described mathematically using the barometric formula, which shows an exponential decrease in pressure with an increase in elevation. The formula can be expressed as: \[ p(z) = p_0 \, e^{-\frac{mgz}{RT}} \] Here, \( p_0 \) is the sea level pressure, \( m \) is the molar mass of air, \( g \) is the acceleration due to gravity, \( z \) is the elevation, \( R \) is the gas constant, and \( T \) is the temperature. To simplify, using standard values like \( g = 9.81 \, \mathrm{m/s^2} \) and \( T = 288.15 \, \mathrm{K} \), we derive that: \[ p(z) = 101.325 \, e^{-0.03416 \, z} \] where \( z \) is in kilometers and \( p(z) \) is in kPa, then convert to atm.

underwater pressure

Underwater pressure increases with depth. Unlike atmospheric pressure, which decreases with elevation, underwater pressure becomes greater as you go deeper. This is due to the weight of the water above. The formula to calculate this pressure is: \[ p(z) = p_0 + \rho_g z \] where \( p_0 \) is the atmospheric pressure at sea level, \( \rho \) is the density of seawater, and \( g \) is the acceleration due to gravity. Using standard values, we can substitute \( \rho = 1046.45 \, \mathrm{kg/m^3} \) and \( g = 9.81 \, \mathrm{m/s^2} \): \[ p(z) = 101.325 + 1046.45 \times 9.81 \times z \] This tells us that the pressure increases linearly with depth.

barometric formula

The barometric formula is used to calculate the change in atmospheric pressure with height. It assumes an exponential decrease in pressure and is based on the ideal gas law and the hydrostatic equation. The standard form of the barometric formula is: \[ p(z) = p_0 \, e^{-\frac{mgz}{RT}} \] To use this formula, you need values for:

• \( p_0 \) (sea level pressure)
• \( m \) (molar mass of air)
• \( g \) (gravity)
• \( z \) (elevation)
• \( R \) (ideal gas constant)
• \( T \) (temperature)

After substituting these values, the formula gives the atmospheric pressure at different elevations.

depth pressure calculation

When calculating the pressure at different depths underwater, it is crucial to understand that pressure increases linearly with depth. The formula used for this calculation is: \[ p(z) = p_0 + \rho g z \] Here, \( \rho \) represents the density of seawater, which is generally constant, and \( g \) is the acceleration due to gravity. Specific values might change with different conditions, but typically for seawater at sea level, you would use: \[ \rho = 1046.45 \, \mathrm{kg/m^3} \] and \[ g = 9.81 \, \mathrm{m/s^2} \]. This ensures that we can predict the increase in pressure accurately as one descends below sea level.

pressure vs. elevation

The relationship between pressure and elevation is inverse. As elevation increases, atmospheric pressure decreases. This means that when you are on top of a mountain, the pressure is lower than at sea level because there is less air above you. This relationship is explained by the barometric formula: \[ p(z) = p_0 \, e^{-\frac{mgz}{RT}} \] Conversely, underwater pressure increases with depth due to the increasing weight of the water above. The deeper you go, the greater the pressure: \[ p(z) = p_0 + \rho g z \]. This direct and linear increase underlines how challenging conditions can be for underwater vehicles as they venture deeper.