On a warm summer day, a large mass of air (atmospheric pressure $1.01 \times 10^{5}$ Pa) is heated by the ground to $26.0$ °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 $g equals 1.40$ . (This rate of cooling for dry, rising air, corresponding to roughly $1$ C° per $100$ m of altitude, is called the dry adiabatic lapse rate.)

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Recognize that the problem involves an adiabatic process, where no heat is exchanged with the surroundings. This is because the air mass rises quickly enough that there is little time for heat exchange, making it approximately adiabatic.

Use the adiabatic process equation for an ideal gas: $ P_1 V_1^\gamma = P_2 V_2^\gamma $, where $ \gamma = 1.40 $ for air. This equation relates the initial and final pressures and volumes of the gas.

Since the process is adiabatic, we can also use the relation between pressure and temperature: $ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}} $. This will allow us to find the final temperature $ T_2 $.

Convert the initial temperature from Celsius to Kelvin: $ T_1 = 26.0 + 273.15 $.

Substitute the known values into the temperature-pressure relation: $ T_2 = T_1 \times \left( \frac{0.850 \times 10^5}{1.01 \times 10^5} \right)^{\frac{1.40 - 1}{1.40}} $. Calculate $ T_2 $ to find the temperature of the air mass at the new pressure.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Adiabatic Process

An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings. In the context of rising air, the process is considered adiabatic because the air mass expands and cools as it rises, without exchanging heat with the cooler surrounding air. This assumption simplifies calculations by focusing on changes in pressure and volume.

Ideal Gas Law

The ideal gas law relates the pressure, volume, and temperature of an ideal gas through the equation PV = nRT, where P is pressure, V is volume, T is temperature, n is the number of moles, and R is the ideal gas constant. This law is crucial for understanding how the temperature of the air mass changes as it rises and the pressure decreases, assuming the air behaves as an ideal gas.

Dry Adiabatic Lapse Rate

The dry adiabatic lapse rate is the rate at which a rising parcel of dry air cools as it expands in the atmosphere, typically about 1°C per 100 meters of altitude. This concept is essential for calculating the temperature change of the air mass as it rises, given that the process is adiabatic and the air is assumed to be dry, meaning it contains no water vapor.

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Textbook Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at $1.00 \times 10^{5}$ Pa and occupies a volume of $2.50 \times 10^{- 3}$ m³. If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

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Textbook Question

A monatomic ideal gas that is initially at $1.50$\times$10^5$ Pa and has a volume of $0.0800$ m³ is compressed adiabatically to a volume of $0.0400$ m³. What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

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Textbook Question

The engine of a Ferrari F355 F1 sports car takes in air at $20.0$ °C and $1.00$ atm and compresses it adiabatically to $0.0900$ times the original volume. The air may be treated as an ideal gas with $g equals 1.40$ . Find the final temperature and pressure.

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Textbook Question

A player bounces a basketball on the floor, compressing it to $80.0 percent sign$ of its original volume. The air (assume it is essentially N₂ gas) inside the ball is originally at $20.0$ °C and $2.00$ atm. The ball's inside diameter is $23.9$ cm. What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal.