Problem 35 - University Physics with Modern Physics

Question

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5}\) Pa) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5}\) Pa. Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cooling for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

Step-by-Step Solution

Answer

The final temperature of the air mass is approximately \(12.22^{\circ}C\).

Step 1: Understanding Adiabatic Process

In an adiabatic process, no heat is exchanged with the environment. Air rising and cooling or descending and warming quickly enough doesn't have time to exchange heat, thus behaving adiabatically.

Step 2: Setting Up the Formula

For adiabatic processes involving ideal gases, the following relation holds: \( P_1 \left(\frac{T_1}{T_2}\right)^\frac{\gamma}{\gamma-1} = P_2 \), where \(P_1\), \(T_1\) are the initial pressure and temperature, and \(P_2\), \(T_2\) are the final pressure and temperature. \(\gamma\) is the heat capacity ratio.

Step 3: Substitute Known Values

Substitute the given values into the adiabatic equation: \( P_1 = 1.01 \times 10^5 \) Pa, \( P_2 = 0.850 \times 10^5 \) Pa, \( T_1 = 26.0 + 273.15 = 299.15 \ K \), and \( \gamma = 1.40 \).

Step 4: Apply the Adiabatic Equation

Using the equation setup, \(1.01 \times 10^5 \left(\frac{299.15}{T_2}\right)^\frac{1.40}{0.40} = 0.850 \times 10^5 \).

Step 5: Solve for Final Temperature

Rearrange and solve the equation: \(\left(\frac{299.15}{T_2}\right)^{3.5} = \frac{0.850}{1.01}\). Solve for \(T_2\) to find the new temperature.

Step 6: Calculation of Temperature

Calculate \(T_2\) by finding the inverse of the power: \(T_2 = 299.15 \times \left(\frac{0.850}{1.01}\right)^{1/3.5}\). Evaluating numerically gives \(T_2 \approx 285.37 \ K\), which is around \(12.22^{\circ}C\).

Key Concepts

Ideal Gas

An ideal gas is a hypothetical concept that helps us understand the behavior of real gases under certain conditions. An ideal gas perfectly follows the ideal gas law, expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin. While no real gas is truly ideal, gases at high temperatures and low pressures tend to behave similarly, thus simplifying calculations with the ideal gas model.

The assumption that air behaves like an ideal gas is useful because it allows us to utilize this simple mathematical relationship in various calculations. This includes understanding how pressure, volume, and temperature relate in dynamic environments, such as rising air masses.

• High Temperature, Low Pressure: Under these conditions, gases have minimal intermolecular forces.
• Predictable Behavior: The relationships between pressure, temperature, and volume are consistent.

While deviations exist, the ideal gas approximation provides a foundational tool for predicting gas behavior, especially in atmospheric science and thermodynamics.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of air in the Earth's atmosphere. It decreases with altitude due to the diminishing weight of the overlying air.

At sea level, this pressure is usually around 101,325 Pa (Pascals), or 1 atm. However, as seen in the provided problem, pressure changes as you ascend, impacting weather and flight dynamics.

• Standard Pressure: At sea level, atmospheric pressure is often standardized at 1 atm, equivalent to approximately \( 1.01 \times 10^5 \) Pa.
• Pressure Decreases with Altitude: As altitude increases, the atmospheric pressure drops, which can affect various phenomena like boiling points and sound propagation.
  
  

Understanding atmospheric pressure is crucial for meteorological studies and aircraft design, as it helps predict how gases within and surrounding us will behave at different altitudes.

Adiabatic Lapse Rate

The adiabatic lapse rate is the rate at which a rising air parcel cools as it expands in the atmosphere without exchanging heat. The term 'adiabatic' implies no heat exchange during the process.

For dry air, the typical temperature drop is about 1°C per 100 meters, which is known as the dry adiabatic lapse rate. This results from the decrease in pressure as the air rises, leading to expansion and cooling.

• Dry Adiabatic Lapse Rate: Approximately 1°C/100m for dry, unsaturated air. The specific value can change slightly with temperature and pressure.
• Importance in Weather Systems: Understanding the lapse rate helps meteorologists predict weather patterns, cloud formation, and storm development.

In the context of the exercise, knowing the adiabatic lapse rate allows us to predict temperature changes of a rising air mass, which significantly impacts weather forecasting and atmospheric science. The concept aids in determining how rapidly temperature changes in elevating air, a crucial factor for pilots and climatologists.