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Chapter 5: Problem 1

A stream of air moves with a speed \(w\). Assume that a mass \(m\) of air is stopped adiabatically by an obstacle. (a) Prove that the rise in temperature of this mass of air is given by $$ \Delta T=\frac{w^{2} M}{5 R} $$ where \(M\) is the molar mass of air. (b) Calculate \(\Delta T\) when \(w=600\) miles \(/ \mathrm{h}\). (c) Apply the equation in part \((a)\) to a meteor moving through a stationary atmosphere at a speed of 20 miles/s. What would happen?

Short Answer

Expert verified
The rise in temperature \Delta T is \frac{w^{2}M}{5R}\. For \w = 600 \/ \text{mi/h}\, \Delta T is calculated using the formula. For the meteor, \Delta T will be extremely high.

Step by step solution

01

Analyze the problem statement

Identify the key information provided: an air stream moving with speed \(w\), a mass \(m\) of air, and the process being adiabatic.
02

Understand adiabatic process

An adiabatic process is one in which no heat is exchanged with surroundings. Hence, the change in internal energy (ndelta Ue) of the system is equal to the work done on or by the system.
03

Apply conservation of energy

For an adiabatic process, kinetic energy is converted to internal energy. \( \frac{1}{2} mv^{2} = \frac{3}{2} n R \Delta T \), where \( \Delta T \) is the rise in temperature.
04

Relate mass and molar quantities

Express the mass \( m \) in terms of moles (n) and the molar mass (M): \( m = nM \), where \( n = \frac{m}{M} \).
05

Derive the expression for \( \Delta T \)

Starting with \( \frac{1}{2} mw^{2} = \frac{3}{2} nR \Delta T \),
06

Solve for \( \Delta T \)

Substitute \( n = \frac{m}{M} \), and simplify to get \[ \Delta T = \frac{w^{2}M}{5R} \].
07

Calculate \Delta T for w=600 miles/hour

Convert speed to consistent units (e.g. meters/second), then plug in values: \[ \Delta T = \frac{(600 \times 0.44704)^{2} \times M}{5R} \].
08

Calculate rise in temperature for a meteor

Convert meteor speed to consistent units (e.g. meters/second) and apply the formula \[ \Delta T = \frac{(20 \times 1609.34)^{2} \times M}{5R} \].

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Key Concepts

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

Adiabatic Process
An adiabatic process is a thermodynamic process in which no heat is exchanged between the system and its surroundings. This can be either because the process happens too rapidly for heat transfer to occur, or because the system is perfectly insulated.
Commonly, adiabatic processes are found in situations like rapid compression or expansion of gases, where these processes occur without time for heat exchange.
The key point is that all the energy involved in the process is used for work done on or by the system or to change its internal energy.
Conservation of Energy
In an adiabatic process, the principle of conservation of energy is crucial. This means that the total energy remains unchanged.
The kinetic energy of the moving air mass is converted into internal energy, causing a rise in temperature. This can be represented by the equation: \(\frac{1}{2} mw^{2} = \frac{3}{2} n R \Delta T\). Here, \(m\) is the mass of air, \(w\) is the speed, \(R\) is the gas constant, and \(\Delta T\) is the temperature rise.
This principle links the initial kinetic energy to the eventual rise in temperature, effectively transforming one form of energy (kinetic) into another (thermal).
Temperature Rise Calculation
To calculate the rise in temperature (\(\Delta T\)) in an adiabatic process, you'll need to know the kinetic energy of the moving mass of air and follow these steps:
1. Identify the given speed \(w\) and mass \(m\).
2. Use the formula: \(\Delta T = \frac{w^{2} M}{5 R}\), where \(M\) is the molar mass of air.
3. Convert any units as needed to keep them consistent, e.g., converting speed from miles per hour to meters per second.
4. Apply the formula with the given values to find the resultant temperature rise.
This step-by-step method ensures that you accurately determine the temperature rise due to the adiabatic stopping of the air mass.
Kinetic Energy Conversion
In an adiabatic process, kinetic energy is fully converted into internal energy. If a mass \(m\) of air moving at speed \(w\) is suddenly stopped, its kinetic energy \(\frac{1}{2}mw^{2}\) transforms entirely into thermal energy.
This full conversion leads to an increase in the temperature of the air.
Knowing that no energy is lost to the surroundings, understanding this conversion is fundamental for solving adiabatic process problems.
This concept underscores why the rise in temperature can be calculated using the kinetic energy formula and why it is important to consider the system as energetically isolated.

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Most popular questions from this chapter

A steel ball of mass \(10 \mathrm{~g}\) is placed in the tube of cross-sectional area \(1 \mathrm{~cm}^{2}\) in Rüchhardt's apparatus. The tube is connected to a jar of air having a capacity of 5 liters, the pressure of the air being \(76 \mathrm{~cm} \mathrm{Hg}\). (a) What is the period of vibration for the ball? (b) If the ball is held initially at a position where the air pressure is exactly atmospheric and then allowed to fall, how far will the ball drop before it starts to come up?

A cylindrical cocktail glass \(15 \mathrm{~cm}\) high and \(35 \mathrm{~cm}^{2}\) in cross section contains water up to the \(10-\mathrm{cm}\) mark. A card is placed over the top and held there while the glass is inverted. When the support for the card is removed, what mass of water must leave the glass in order that the rest of the water will remain in the glass, if one neglects the weight of the card? (Caution: Try this over a sink.)

A horizontal, insulated cylinder contains a frictionless nonconducting piston. On each side of the piston is 54 liters of an inert monatomic ideal gas at 1 atm and \(273 \mathrm{~K}\). Heat is slowly supplied to the gas on the left side until the piston has compressed the gas on the right side to \(7.59 \mathrm{~atm}\). (a) How much work is done on the gas on the right side? (b) What is the final temperature of the gas on the right side? (c) What is the final temperature of the gas on the left side? (d) How much heat was added to the gas on the left side?

(a) Derive the following formula for a quasi-static adiabatic process for the ideal gas, assuming \(\gamma\) to be constant: $$ \frac{T}{P(\gamma-1) / \gamma}=\text { const. } $$ (b) Helium \(\left(\gamma=\frac{5}{3}\right)\) at \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}\) pressure is compressed quasi-statically and adiabatically to a pressure of 5 atm. Assuming that the helium behaves like the ideal gas, calculate the final temperature.

A thick-walled insulated chamber contains \(n_{i}\) moles of helium at high pressure \(P_{i}\). It is connected through a valve with a large, almost empty container of helium at constant pressure \(P_{0}\), very nearly atmospheric. The valve is opened slightly, and the helium flows slowly and adiabatically into the container until the pressures on the two sides of the valve are equal. Assuming the helium to behave like an ideal gas with constant heat capacities, show that: (a) The final temperature of the gas in the chamber is $$ T_{f}=T_{i}\left(\frac{P_{f}}{P_{i}}\right)^{(\gamma-1) / \gamma} $$ (b) The number of moles left in the chamber is $$ n_{f}=n_{i}\left(\frac{P_{f}}{P_{i}}\right)^{1 / \gamma} \text { . } $$ (c) The final temperature of the gas in the container is $$ T_{f}=\frac{T_{i}}{\gamma} \frac{1-P_{f} / P_{i}}{1-\left(P_{f} / P_{i}\right)^{1 / \gamma}} . $$
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