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It has been suggested that air at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) can be cooled by first compressing it adiabatically in a closed system to \(1000 \mathrm{kPa}\) and then expanding it adiabatically back to \(100 \mathrm{kPa} .\) Is this possible?

Short Answer

Expert verified
Answer: No, the process of adiabatic compression followed by adiabatic expansion does not result in the cooling of the air, as the initial and final temperatures are the same.

Step by step solution

01

Find the initial and final temperatures (in Kelvin)

Convert the given initial temperature from Celsius to Kelvin. Initial temperature \(T_1 = 25^\circ \mathrm{C} + 273.15 = 298.15\,\mathrm{K}\). Since the final pressure after expansion is the same as the initial pressure (\(100\,\mathrm{kPa}\)), we need to find the temperature after each stage of the process.
02

Set up the ideal gas law for adiabatic compression

An adiabatic process satisfies the relationship: \(T_1P_1^{\frac{\gamma - 1}{\gamma}} = T_2P_2^{\frac{\gamma - 1}{\gamma}}\) Where \(\gamma\) is the heat capacity ratio (specific heat at constant pressure divided by specific heat at constant volume). For air, \(\gamma = 1.4\). We can rearrange and solve for \(T_2\), the temperature after adiabatic compression.
03

Calculate the temperature after adiabatic compression

Given \(P_1 = 100\,\mathrm{kPa}\), \(P_2 = 1000\, \mathrm{kPa}\), and initial temperature \(T_1 = 298.15\,\mathrm{K}\). We can now plug these values into the equation from Step 2: \(T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} = 298.15 \left(\frac{1000}{100}\right)^{\frac{1.4 - 1}{1.4}}\) Calculating this, we get: \(T_2 \approx 536.94\, \mathrm{K}\)
04

Set up the ideal gas law for adiabatic expansion

Now, let's consider the adiabatic expansion back to the initial pressure. We can use the same adiabatic process equation: \(T_2P_2^{\frac{\gamma - 1}{\gamma}} = T_3P_1^{\frac{\gamma - 1}{\gamma}}\) Where \(T_3\) is the temperature after adiabatic expansion.
05

Calculate the temperature after adiabatic expansion

Re-arrange and compute for \(T_3\): \(T_3 = T_2 \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}} = 536.94 \left(\frac{100}{1000}\right)^{\frac{1.4 - 1}{1.4}}\) Calculating, we obtain: \(T_3 \approx 298.15\, \mathrm{K}\)
06

Compare the initial and final temperatures

Now, compare the initial temperature \(T_1 = 298.15\, \mathrm{K}\) to the final temperature \(T_3 \approx 298.15\, \mathrm{K}\). We can see that the initial and final temperatures are the same. Thus, the suggested process of adiabatic compression followed by adiabatic expansion does not result in the cooling of the air. The answer to "Is this possible?" is No.

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Key Concepts

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

Adiabatic Compression
Adiabatic compression is a significant concept in thermodynamics, particularly in the context of heat engine cycles and atmospheric phenomena. Fundamentally, during adiabatic compression, a gas is compressed, leading to a rise in pressure and temperature without any exchange of heat with the surroundings. This is often demonstrated in the real world when you pump air into a bicycle tire; the pump gets warm as a result of adiabatic compression of the air.

The process is governed by the principle of energy conservation, which states that when work is done on the gas to compress it, its internal energy increases, thereby raising its temperature. In an idealized scenario, the absence of heat transfer means that all of the work done on the gas directly increases its internal energy.
Ideal Gas Law
The ideal gas law is a cornerstone in the study of thermodynamics and an invaluable tool for scientists and engineers. It relates the pressure (P), volume (V), and temperature (T) of an ideal gas to the amount of gas present (n), usually measured in moles, and a universal constant (R), known as the ideal gas constant. Expressed in the form of an equation, it is written as:

\[ PV = nRT \]
This equation assumes that the gas particles do not interact with each other and that their size is negligible compared to the distance between them. While this model is an approximation, it offers a remarkably accurate description of the behavior of real gases under many conditions. The ideal gas law is especially helpful for analyzing adiabatic processes because it helps us determine the state of a gas before and after the process.
Heat Capacity Ratio
The heat capacity ratio, often represented by the Greek letter gamma (\(\gamma\)), is a dimensionless quantity that plays a pivotal role in characterizing adiabatic processes. It is defined as the ratio of the specific heat capacity at constant pressure (\(C_p\)) to the specific heat capacity at constant volume (\(C_v\)):

\[ \gamma = \frac{C_p}{C_v} \]
For diatomic gases, like oxygen and nitrogen, which make up the majority of air, the heat capacity ratio is approximately 1.4. This value is crucial in predicting how the temperature of a gas changes with pressure during adiabatic processes without the need for detailed calculations of the actual heat capacities. In adiabatic compression or expansion, it allows us to use the relation:

\[ T_1P_1^{\frac{\gamma - 1}{\gamma}} = T_2P_2^{\frac{\gamma - 1}{\gamma}} \]
This equation forms the backbone of analyses for processes such as the one described in the exercise.

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Most popular questions from this chapter

Liquid water enters an adiabatic piping system at \(15^{\circ} \mathrm{C}\) at a rate of \(8 \mathrm{kg} / \mathrm{s} .\) If the water temperature rises by \(0.2^{\circ} \mathrm{C}\) during flow due to friction, the rate of entropy generation in the pipe is \((a) 23 \mathrm{W} / \mathrm{K}\) \((b) 55 \mathrm{W} / \mathrm{K}\) \((c) 68 \mathrm{W} / \mathrm{K}\) \((d) 220 \mathrm{W} / \mathrm{K}\) \((e) 443 \mathrm{W} / \mathrm{K}\)

A well-insulated heat exchanger is to heat water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 25 to \(60^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{kg} / \mathrm{s}\) The heating is to be accomplished by geothermal water \(\left(c_{p}=4.31 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) available at \(140^{\circ} \mathrm{C}\) at a mass flow rate of \(0.75 \mathrm{kg} / \mathrm{s} .\) Determine \((a)\) the rate of heat transfer and \((b)\) the rate of entropy generation in the heat exchanger.

You are to expand a gas adiabatically from 3 MPa and \(300^{\circ} \mathrm{C}\) to \(80 \mathrm{kPa}\) in a piston-cylinder device. Which of the two choices - air with an isentropic expansion efficiency of 90 percent or neon with an isentropic expansion efficiency of 80 percent - will produce the most work?

Steam expands in an adiabatic turbine from \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) to \(0.1 \mathrm{MPa}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\). If steam leaves the turbine as saturated vapor, the power output of the turbine is \((a) 2058 \mathrm{kW}\) \((b) 1910 \mathrm{kW}\) \((c) 1780 \mathrm{kW}\) \((d) 1674 \mathrm{kW}\) \((e) 1542 \mathrm{kW}\)

An insulated piston-cylinder device initially contains \(0.02 \mathrm{m}^{3}\) of saturated liquid-vapor mixture of water with a quality of 0.1 at \(100^{\circ} \mathrm{C} .\) Now some ice at \(-18^{\circ} \mathrm{C}\) is dropped into the cylinder. If the cylinder contains saturated liquid at \(100^{\circ} \mathrm{C}\) when thermal equilibrium is established, determine (a) the amount of ice added and ( \(b\) ) the entropy generation during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\).

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