Problem 38 - University Physics with Modern Physics

Question

On a warm summer day, a large mass of air (armospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) 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 \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of attitude, is called the dry adiabatic lapse rate.)

Step-by-Step Solution

Answer

The temperature of the air mass is approximately \(12.17^{\circ} \mathrm{C}\) at the new pressure.

Step 1: Understand the Adiabatic Process

An adiabatic process is where no heat is exchanged with the surroundings. For rising air, this applies because the expansion due to decreased pressure results in cooling without heat loss to the cooler air.

Step 2: Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin. \( 26.0^{\circ} \mathrm{C} = 26.0 + 273.15 = 299.15 \ \mathrm{K} \).

Step 3: Use the Adiabatic Relation for Ideal Gases

The adiabatic relation is \( T_1 P_1^{\gamma-1} = T_2 P_2^{\gamma-1} \). We know the initial temperature \( T_1 = 299.15 \ \mathrm{K} \), initial pressure \( P_1 = 1.01 \times 10^5 \ \mathrm{Pa} \), final pressure \( P_2 = 0.850 \times 10^5 \ \mathrm{Pa} \), and \( \gamma = 1.40 \). We must find \( T_2 \).

Step 4: Rearrange and Solve the Equation

Rearrange the adiabatic equation to solve for \( T_2 \): \[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \].

Step 5: Calculate the Temperature at New Pressure

Substitute the known values into the equation: \[ T_2 = 299.15 \left( \frac{0.850 \times 10^5}{1.01 \times 10^5} \right)^{\frac{1.40-1}{1.40}} \].

Step 6: Simplify and Calculate

Calculate the ratio \( \frac{0.850}{1.01} \approx 0.841584 \). Then compute \( 0.841584^{0.286} \approx 0.954 \). Finally, calculate \( T_2 = 299.15 \times 0.954 \approx 285.32 \ \mathrm{K} \).

Step 7: Convert Temperature Back to Celsius

Convert the final temperature back to Celsius: \( 285.32 \ \mathrm{K} - 273.15 = 12.17^{\circ} \mathrm{C} \).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle used to describe the behavior of gases under various conditions. It is represented by the equation \( PV = nRT \), where \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. This law assumes that the gas behaves ideally, meaning:

• The particles do not attract or repel each other.
• The particles themselves take up negligible volume compared to the container.
• The particles are in random motion and collisions are perfectly elastic.

For our exercise with the rising air mass, although the Ideal Gas Law itself isn't directly applied in calculating the temperature, understanding that air behaves approximately as an ideal gas helps us utilize the adiabatic process relations effectively when pressure and temperature change but the amount of gas doesn't.

Adiabatic Lapse Rate

The adiabatic lapse rate is the rate of temperature change experienced by a mass of air as it moves upward and experiences pressure changes adiabatically (without heat exchange with the environment). There are two types:

• Dry Adiabatic Lapse Rate (DALR): Typically around \(1^\circ \mathrm{C} \) per 100 meters for dry, uprising air. This occurs when no moisture condensation takes place.
• Moist Adiabatic Lapse Rate (MALR): Less than DALR due to latent heat release when water vapor condenses.

In the problem, since the gypsum process is dry, we use the dry adiabatic lapse rate. As air rises, it expands and cools because the pressure decreases. Our given conditions illustrate this concept: as pressure drops from \(1.01 \times 10^5 \mathrm{Pa} \) to \(0.850 \times 10^5 \mathrm{Pa} \), temperature similarly drops from \(26.0^\circ \mathrm{C}\) to about \(12.17^\circ \mathrm{C} \). This cooling is an adiabatic change, as heat is not exchanged with the outside air.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and the forms of energy transformation. In the context of this problem, we focus on adiabatic processes, a key area of thermodynamics:

• An adiabatic process is a type of thermodynamic process that occurs without exchanging heat between the system and its surroundings.
• It is characterized by changes in pressure, temperature, and volume where internal energy changes are solely due to work done by or on the system.
• The basic formula for adiabatic processes in ideal gases is \( P_1V_1^\gamma = P_2V_2^\gamma \) or alternatively \( T_1P_1^{\gamma-1} = T_2P_2^{\gamma-1} \).

For the air mass rising scenario, understanding these principles of thermodynamics allows us to deduce how pressure and temperature are related during the air's ascent. The equation adjustments from the exercise further illustrate the adiabatic behavior: cooling occurs through expansion as the air rises, without heat loss.