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Chapter 19: Problem 31

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5} \mathrm{~Pa}\) ) is heated by the ground to \(26.0^{\circ} \mathrm{C}\) 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 cooling for dry, rising air, corresponding to roughly \(1 \mathrm{C}^{\circ}\) per \(100 \mathrm{~m}\) of altitude, is called the \(d r y\) adiabatic lapse rate.

Short Answer

Expert verified
The final temperature of the air mass when it has risen to the level where the atmospheric pressure is \(0.850 \times 10^{5} ~\mathrm{Pa}\) is calculated using an adiabatic process formula given the initial conditions of the air mass, the ratio of specific heats for the ideal gas and the final pressure. The result in Kelvin is then converted to Celsius for the final answer.

Step by step solution

01

Note down Given Values and Convert Temperature to Kelvin

Given initial pressure \(P_i = 1.01 \times 10^{5} ~\mathrm{Pa}\), initial temperature \(T_i = 26.0^{\circ} \mathrm{C}\), and final pressure \(P_f = 0.850 \times 10^{5}\). Convert initial temperature \(T_i\) to Kelvin by adding 273.15 to the Celsius value, thus \(T_i = 26.0 + 273.15 = 299.15 ~\mathrm{K}\). The ratio of specific heats \(\gamma = 1.4\). The final temperature \(T_f\) is to be determined.
02

Apply the Law of Adiabatic Processes

Adiabatic processes follow the formula \(T_iP_i^{(\gamma - 1)/\gamma} = T_fP_f^{(\gamma - 1)/\gamma}\). Substituting the given values, we have \(299.15 \times (1.01 \times 10^{5})^{(\gamma - 1)/\gamma} = T_f \times (0.850 \times 10^{5})^{(\gamma - 1)/\gamma}\). This equation can be solved for \(T_f\).
03

Solve for the final temperature

Rearrange the equation from step 2 and solve for \(T_f\), yielding \(T_f = \frac{299.15 \times (1.01 \times 10^{5})^{1-1/\gamma}}{(0.850 \times 10^{5})^{1-1/\gamma}}\). Plugging in the value for \(\gamma\), this provides the temperature of the air mass \(T_f\) when it has risen to the level at which atmospheric pressure is \(0.850 \times 10^{5} ~\mathrm{Pa}\). Note that the answer will also be in Kelvin, Convert the result from Kelvin to Celsius by subtracting 273.15.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that links the pressure, volume, temperature, and amount of gas in a system. Expressed as PV=nRT, where P represents the pressure, V the volume, n the amount of substance in moles, R the ideal gas constant, and T the temperature in Kelvin. This equation assumes that the gas is ideal, meaning it follows certain theoretical behaviors such as elastic collisions and negligible volume compared to the container.

Understanding the ideal gas law is essential for solving problems related to adiabatic processes because it allows us to predict how a gas's pressure and temperature will change without the exchange of heat. In the example exercise, the ideal gas law helps in understanding why the temperature of the rising air mass decreases as the pressure likewise drops, without the need to consider the exchange of heat with surrounding air.
Specific Heat Ratio

Understanding \(\gamma\)

The specific heat ratio, often denoted as \(\gamma\), is the ratio of the specific heat at constant pressure (\(C_p\)) to the specific heat at constant volume (\(C_v\)). This dimensionless quantity is particular for each gas and influences how the gas will undergo thermodynamic processes.

For instance, in diatomic gases such as nitrogen and oxygen, which make up the majority of Earth's atmosphere, \(\gamma \'s\) value is typically around 1.40. This figure is pivotal when applying the adiabatic process formula, as seen in the exercise, allowing us to relate the initial and final states of pressure and temperature for a rising or descending air parcel. A higher \(\gamma\) indicates a gas that will undergo greater temperature changes during adiabatic processes, compared to a gas with a lower \(\gamma\).
Dry Adiabatic Lapse Rate
The dry adiabatic lapse rate (DALR) is the rate at which a dry parcel of air cools as it rises or warms as it descends, assuming no heat is transferred to or from the environment鈥攈ence the term 'adiabatic'. It's a product of the specific heat ratio and the force of gravity over the specific gas constant for dry air.

The DALR is typically about \(9.8 ^\circ C/1000 \'m\) and it is a key concept in meteorology because it predicts the cooling of an air mass as it rises through the atmosphere. If the surrounding atmosphere is cooler than the DALR, the rising air will continue to be buoyant; if it's warmer, the air will tend to stop rising. This rate is crucial for understanding weather patterns, cloud formation, and the stability of the atmosphere. The exercise provided is a practical application of the DALR, as it requires calculating the temperature change of an air mass based on its change in altitude determined by pressure changes.

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Most popular questions from this chapter

DATA You compress a gas in an insulated cylinder-no heat flows into or out of the gas. The gas pressure is fairly low, so treating the gas as ideal is a good approximation. When you measure the pressure as a function of the volume of the gas, you obtain these results: $$ \begin{array}{l|lllll} \boldsymbol{V}(\mathbf{L}) & 2.50 & 2.02 & 1.48 & 1.01 & 0.50 \\ \hline \boldsymbol{p}(\mathbf{a t m}) & 0.101 & 0.139 & 0.202 & 0.361 & 0.952 \end{array} $$ (a) Graph \(\log (p)\) versus \(\log (V),\) with \(p\) in \(\mathrm{Pa}\) and \(V\) in \(\mathrm{m}^{3} .\) Explain why the data points fall close to a straight line. (b) Use your graph to calculate \(\gamma\) for the gas. Is the gas monatomic, diatomic, or polyatomic? (c) When \(p=0.101\) atm and \(V=2.50 \mathrm{~L}\), the temperature is \(22.0^{\circ} \mathrm{C}\). Apply the ideal-gas equation and calculate the temperature for each of the other pairs of \(p\) and \(V\) values. In this compression, does the temperature of the gas increase, decrease, or stay constant?

19.52 - A certain ideal gas has molar heat capacity at constant volume \(C_{V}\). A sample of this gas initially occupies a volume \(V_{0}\) at pressure \(p_{0}\) and absolute temperature \(T_{0}\). The gas expands isobarically to a volume \(2 V_{0}\) and then expands further adiabatically to a final volume \(4 V_{0}\). (a) Draw a \(p V\) -diagram for this sequence of processes. (b) Compute the total work done by the gas for this sequence of processes. (c) Find the final temperature of the gas. (d) Find the absolute value of the total heat flow \(|Q|\) into or out of the gas for this sequence of processes, and state the direction of heat flow.

A cylinder contains \(0.250 \mathrm{~mol}\) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\). The cylinder is provided with a friction less piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\). Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

During an adiabatic expansion the temperature of \(0.450 \mathrm{~mol}\) of argon (Ar) drops from \(66.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\). The argon may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

\( \mathrm{A}\) gas in a cylinder expands from a volume of \(0.110 \mathrm{~m}^{3}\) to \(0.320 \mathrm{~m}^{3} .\) Heat flows into the gas just rapidly enough to keep the pres- sure constant at \(1.65 \times 10^{5} \mathrm{~Pa}\) during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{~J}\). (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?
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