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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\(^\circ\)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 \(\Upsilon\) = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C\(^\circ\) per 100 m of altitude, is called the dry \(adiabatic\) \(lapse\) \(rate\).)

Short Answer

Expert verified
T ≈ 20.4°C

Step by step solution

01

Understanding the Adiabatic Process

The process is adiabatic because the air mass does not exchange heat with its surroundings as it rises, due to the rapid nature of the expansion. In an adiabatic process, the change in internal energy is equal to the work done on or by the system.
02

Apply the Adiabatic Equation

For adiabatic processes involving an ideal gas, the equation \( P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \) and the adiabatic relation \( \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \) can be used. Here, \( \gamma \) is the heat capacity ratio, given as 1.40.
03

Calculate Initial Parameters

Set the initial temperature \( T_1 = 26.0^{\circ}C = 299.15 K \). The initial pressure \( P_1 = 1.01 \times 10^5 \) Pa, and the final pressure \( P_2 = 0.850 \times 10^5 \) Pa.
04

Use the Adiabatic Relation to Find Final Temperature

Substitute into the adiabatic relation: \( T_2 = T_1 \times \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} \). Calculate \( T_2 \):\[T_2 = 299.15 \times \left( \frac{0.850 \times 10^5}{1.01 \times 10^5} \right)^{\frac{1.40-1}{1.40}}\]Evaluate to find \( T_2 \).
05

Substitute and Solve the Expression

Perform the computation:\[T_2 = 299.15 \times \left( \frac{0.850}{1.01} \right)^{0.286}\]Evaluate the powers and multiplication to find the final temperature in Kelvin.
06

Convert Temperature Back to Celsius

After calculating the final temperature \( T_2 \) in Kelvin, convert it back to degrees Celsius using the relation \( T(^{\circ}C) = T(K) - 273.15 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas
An ideal gas is a theoretical concept that simplifies the complex behavior of gases in nature. It is assumed to follow certain principles that make calculations easier. Primarily, an ideal gas obeys the Ideal Gas Law, which states that the product of pressure and volume is proportional to the temperature.
At a microscopic level, ideal gas particles are considered as point particles that do not interact, except through elastic collisions.
  • They are assumed to have no volume.
  • No intermolecular forces exist between them.
  • Collisions are perfectly elastic, meaning no energy is lost.
Despite not existing in reality, the concept of an ideal gas allows us to make accurate approximations under many conditions, especially when the gas is at high temperature and low pressure. These conditions minimize interactions that deviate from ideal behavior.
Dry Adiabatic Lapse Rate
When a parcel of air rises, it expands and cools due to the lower pressure at higher altitudes. If the process occurs without heat exchange with the environment, it is termed adiabatic. In this context, the dry adiabatic lapse rate (DALR) describes how the temperature of dry air changes as it moves vertically.
The typical dry adiabatic lapse rate is about 1°C per 100 meters. It's a critical concept for meteorology, as it helps predict temperature changes in rising air masses. The DALR is calculated considering the following principles:
  • The atmosphere's pressure decrease with altitude causes expansion.
  • No heat exchange, hence adiabatic.
  • Dry air, meaning no condensation takes place.
The significance of the DALR is crucial for weather forecasting and understanding cloud formation.
Heat Capacity Ratio
The heat capacity ratio, also known as the adiabatic index or gamma (\(\gamma\)), is the ratio of specific heat at constant pressure (\(C_p\)) to specific heat at constant volume (\(C_v\)). In the context of adiabatic processes, it plays a vital role in determining how the temperature of an ideal gas changes with pressure.
  • For air, \(\gamma\) is typically around 1.40.
  • It reflects how much energy is used in expansion versus heating.
  • The value of the heat capacity ratio varies for different gases.
In an adiabatic process, knowing the heat capacity ratio is essential because it helps in calculating how the temperature of a gas will change without adding or extracting heat. Mathematically, it appears in the adiabatic relation \(\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}\). Hence, it's crucial for engineers and atmospheric scientists alike to understand and apply \(\gamma\) correctly in real-world situations.

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Most popular questions from this chapter

Starting with 2.50 mol of N\(_2\) gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0\(^\circ\)C, a chemist first heats the gas at constant volume, adding 1.36 \(\times\) 10\(^4\) J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 \(\times\) 10\(^5\) Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(pV\)-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

A gas in a cylinder expands from a volume of 0.110 m\(^3\) to 0.320 m\(^3\). Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 \(\times\) 10\(^5\) Pa during the expansion. The total heat added is 1.15 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

Five moles of monatomic ideal gas have initial pressure 2.50 \(\times\) 10\(^3\) Pa and initial volume 2.10 m\(^3\). While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?

In an experiment to simulate conditions inside an automobile engine, 0.185 mol of air at 780 K and 3.00 \(\times\) 10\(^6\) Pa is contained in a cylinder of volume 40.0 cm\(^3\). Then 645 J of heat is transferred to the cylinder. (a) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, and use the data in Table 19.1 even though the pressure is not low. Draw a \(pV\)-diagram for this process. (b) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, repeat part (a).

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