Problem 21 - University Physics with Modern Physics

Question

At an altitude of \(11,000 \mathrm{m}\) (a typical cruising altitude for a jet airliner), the air temperature is \(-56.5^{\circ} \mathrm{C}\) and the air density is 0.364 \(\mathrm{kg} / \mathrm{m}^{3} .\) What is the pressure of the atmosphere at that altitude? (Note: The temperature at this altitude is not the same as at the surface of the earth, so the calculation of Example 18.4 in Section 18.1 doesn't apply.)

Step-by-Step Solution

Answer

The atmospheric pressure at 11,000 m is approximately 22641 Pa.

Step 1: Identify the Known Values

From the problem, we know the air temperature \( T \) is \(-56.5^{\circ}C \). We must first convert this to Kelvin: \( T = -56.5^{\circ}C + 273.15 = 216.65 \: K \). The air density \( \rho \) is given as 0.364 kg/m³.

Step 2: Recall the Ideal Gas Law

The Ideal Gas Law is given by \( P = \rho R T \), where \( P \) is the pressure, \( \rho \) is the density, \( R \) is the specific gas constant for dry air, and \( T \) is the temperature in Kelvin. For dry air, \( R = 287 \: J/kg \, K \).

Step 3: Substitute Known Values

Plug the known values into the Ideal Gas Law equation:\[P = 0.364 \, \text{kg/m}^3 \times 287 \, \text{J/(kg} \, \text{K)} \times 216.65 \, \text{K}\]

Step 4: Calculate the Pressure

Upon calculating the expression, we find:\[P = 0.364 \times 287 \times 216.65 = 22640.628 \, \text{Pa}\]Thus, the pressure at that altitude is approximately 22641 Pa.

Key Concepts

Air Temperature and Density

Understanding how temperature and density interact is crucial when dealing with gases. At high altitudes, like 11,000 meters above sea level, the air temperature can be significantly lower than at the Earth's surface. In the exercise, the air temperature is given as -56.5°C, which equates to 216.65 Kelvin when converted.
The conversion from Celsius to Kelvin is simple yet essential for calculations involving gas laws, as these laws typically use Kelvin, the absolute temperature scale. The equation to remember is Kelvin = Celsius + 273.15.
Air density at such an altitude is also much lower, given here as 0.364 kg/m³. This lower density is because fewer air molecules are present at higher altitudes.

• Lower temperatures generally result in increased density under constant pressure, but at the actual atmospheric conditions of high altitude, lower temperatures coincide with lower density due to expansion and mixing of warm and cold air layers.
• Low air density means there are fewer air molecules to exert pressure, which is a key aspect when calculating atmospheric pressure using the ideal gas law.

Pressure Calculation

Determining the atmospheric pressure at high altitudes involves the ideal gas law. This law relates pressure (\( P \)), density (\( \rho \)), the specific gas constant (\( R \)), and temperature (\( T \)).
The formula is:\[ P = \rho R T \]Given that we've converted the temperature to 216.65 \( K \) and already know the air density is 0.364 \( \text{kg/m}^3 \), we can easily substitute these into the formula.

• First, recall that the specific gas constant \( R \) for dry air is 287 \( \text{J/(kg} \, \text{K}) \), as this is a fundamental constant used in these calculations.
• Next, substitute the values into the ideal gas equation: \( P = 0.364 \times 287 \times 216.65 \).
• Solving this yields \( P = 22640.628 \text{ Pa} \), which rounds to 22641 Pa.

This pressure value shows how thin and less pressurized the air is at 11,000 meters compared to sea level.

Gas Constant for Dry Air

The specific gas constant for dry air, \( R \), is a critical value when using the ideal gas law to calculate atmospheric properties. It defines the relationship between pressure, volume, and temperature for dry air.
The constant, 287 \( \text{J/(kg} \, \text{K}) \), explains how much energy is contained in a mole of gas at a given temperature. This value is derived from the more general gas constant, \( R \), for all gases, divided by the molar mass of dry air.

• The gas constant for a specific gas, such as dry air, allows us to make accurate predictions about how air behaves under changes in temperature and pressure.
• For the exercise, knowing \( R \) enables us to link the physical properties like density and temperature directly to calculate the atmospheric pressure.

Understanding and utilizing this constant allows students to solve real-world problems involving atmospheric conditions at different altitudes.