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Chapter 19: Problem 61

An air pump has a cylinder 0.250 \(\mathrm{m}\) long with a mov- able piston. The pump is used to compress air from the atmosphere (at absolute pressure \(1.01 \times 10^{5}\) Pa) into a very large tank at \(4.20 \times 10^{5}\) Pa gauge pressure. (For air, \(C_{V}=20.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at \(27.0^{\circ} \mathrm{C}\) , what is the temperature of the compressed air? (c) How much work does the pump do in putting 20.0 mol of air into the tank?

Short Answer

Expert verified
(a) 0.179 m; (b) 533 K; (c) 96632 J.

Step by step solution

01

Understanding Adiabatic Compression

In adiabatic compression, no heat exchange occurs between the system and its environment. This process is governed by the relation \( PV^{\gamma} = \text{constant} \), where \( \gamma = \frac{C_P}{C_V} \). For air \( \gamma \) is approximately 1.4.
02

Calculating Final Pressure

The gauge pressure in the tank is given as \(4.20 \times 10^{5}\text{ Pa}\), but gauge pressure does not include atmospheric pressure. The absolute pressure is therefore \(4.20 \times 10^{5} + 1.01 \times 10^5 = 5.21 \times 10^5 \text{ Pa} \).
03

Using Adiabatic Relation to Find Volume Change

Given that \( P_1V_1^{\gamma} = P_2V_2^{\gamma} \), where \( P_1 = 1.01 \times 10^5 \text{ Pa} \), \( P_2 = 5.21 \times 10^5 \text{ Pa} \), and \( \gamma = 1.4 \), set \( V_1 = A \times 0.250 \text{ m} \). Solve for \( V_2 \). You'll find \( V_2 = A \times x \) where \( x \) is the distance the piston has moved.
04

Calculating Piston Displacement

Substitute known values to find \( x \): \( (1.01 \times 10^5 \times (0.250)^{1.4}) = (5.21 \times 10^5 \times x^{1.4}) \), and solve for \( x \). The formula reduces to \( x \approx 0.179 \text{ m} \).
05

Calculating Compressed Air Temperature

For an adiabatic process, \( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \). Knowing \( T_1 = 27^{\circ} \mathrm{C} = 300 \mathrm{K} \), calculate \( T_2 \) with \( P_2 = 5.21 \times 10^5 \text{ Pa} \) and \( P_1 = 1.01 \times 10^5 \text{ Pa} \). You find \( T_2 \approx 533 \text{ K} \).
06

Calculating Work Done by the Pump

The work done \( W \) in an adiabatic process is \( W = nC_V(T_2 - T_1) \). Here, \( n = 20.0 \) moles and \( C_V = 20.8 \text{ J/mol} \cdot \text{K} \). Using \( T_2 = 533 \text{ K} \) and \( T_1 = 300 \text{ K} \), calculate \( W = 20.0 \times 20.8 \times (533 - 300) = 96632 \text{ J} \).

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It is crucial for understanding processes like adiabatic compression, as seen in the problem with the air pump. In an adiabatic process, there is no heat exchange with the environment. This means that all the energy used to compress the gas is transformed into internal energy of the gas, raising its temperature.

Adiabatic Process Basics:
  • Involves no heat exchange (Qt = 0).
  • The energy change is done entirely through work.
  • Commonly found in rapidly occurring or insulated systems.
When we talk about adiabatic processes in thermodynamics, we often use the equation \( PV^\gamma = \text{constant} \), which represents the relationship between pressure (\( P \)), volume (\( V \)), and the adiabatic index, \( \gamma \). Knowing how these variables interact is key to calculating changes in an adiabatic system.
Gas Laws
Gas laws are fundamental to understanding how gases behave under different conditions of pressure, temperature, and volume. Several laws, such as Boyle's Law, Charles's Law, and the Ideal Gas Law, help describe these behaviors. For adiabatic processes and changes in a gas system, the specific heat capacities, \( C_P \) and \( C_V \), play an integral role, with \( \gamma = \frac{C_P}{C_V} \).

Equation for Adiabatic Processes:
  • Boyle's Law: For a constant temperature, \( PV = \text{constant} \).
  • Adiabatic Condition: \( PV^\gamma = \text{constant} \), indicating volume and pressure changes without external heat exchange.
Understanding these principles helps explain why an air pump's piston displacement is determined by the changes in pressure and volume, resulting from compressing the gas without heat exchange, a core aspect of adiabatic compression.

For the given problem, the specific heat at constant volume \( C_V \) also assists in determining the work required to compress the gas, highlighting the importance of gas laws in thermodynamics.
Work and Energy
Work and energy in terms of gas compression are central to solving problems involving adiabatic processes. The work done on a gas results in changes in its internal energy, affecting its temperature and pressure. For the air pump, the work done during compression can be calculated using the first law of thermodynamics, which relates changes in internal energy to the work done on the system and any heat exchange, which is zero in adiabatic compression.

Work Done in Adiabatic Processes:
  • The formula to calculate work done: \( W = nC_V(T_2 - T_1) \).
  • \( W \) symbolizes the work, \( n \) the moles of gas, \( C_V \) the specific heat at constant volume, and \( T_2 \), \( T_1 \) the final and initial temperatures.
This formula helps calculate the total energy consumed during the process, confirming how much effort is needed to compress the gas from one state to another.

This work is why the compressed air's temperature increases after compression. By understanding how energy transfers occur in adiabatic systems, students can better predict outcomes and solve related thermodynamic problems.

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Most popular questions from this chapter

A player bounces a basketball on the floor, compressing it to 80.0\(\%\) of its original volume. The air (assume it is essentially \(\mathrm{N}_{2}\) gas) inside the ball is originally at a temperature of \(20.0^{\circ} \mathrm{C}\) and a pressure of 2.00 atm. The ball's inside diameter is 23.9 \(\mathrm{cm}\) . (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

CALC Acylinder with a piston contains 0.250 mol of \(0 x y-\) gen at \(2.40 \times 10^{5}\) Pa and 355 \(\mathrm{K}\) . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a \(p V\) -diagram. (b) Compute the temperature during the isothermal compression. (c) Compute the maximum pressure. (d) Compute the total work done by the piston on the gas during the series of processes.

Acylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) . (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C} ?\) Draw a \(p V\) -diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

Heat \(Q\) flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is \(2.20 \times 10^{6} \mathrm{J} / \mathrm{kg}\) and the boiling point is \(120^{\circ} \mathrm{C}\) . At this pressure, 1.00 \(\mathrm{kg}\) of water has a volume of \(1.00 \times 10^{-3} \mathrm{m}^{3},\) and 1.00 \(\mathrm{kg}\) of steam has a volume of 0.824 \(\mathrm{m}^{3} .\) (a) Compute the work done when 1.00 \(\mathrm{kg}\) of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.
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