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In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.) (a) What is the final volume of this air after compression? (b) How much work is done in compressing the air? (c) If the temperature of the air is initially \(300 \mathrm{K},\) what is the temperature after compression?

Short Answer

Expert verified
(a) Final volume is approx. 0.349 liters. (b) Work done is approx. 364.68 J. (c) Final temperature is approx. 442.14 K.

Step by step solution

01

Identify Given Information and Assumptions

Given: Initial volume \(V_1 = 1\) liter, initial pressure \(P_1 = 1\) atm, final pressure \(P_2 = 7\) atm. We assume adiabatic compression, which means no heat is exchanged. For diatomic gases, the adiabatic index \( \gamma = \frac{7}{5} = 1.4\).
02

Calculate Final Volume Using Adiabatic Equation

Use the adiabatic process relationship: \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\). Substitute the known values: \[1 \times (1)^{{1.4}} = 7 \times V_2^{1.4}\] Solving for \(V_2\), \[V_2 = \left(\frac{1}{7}\right)^{\frac{1}{1.4}}\] \[V_2 \approx 0.3486\text{ liters}\].
03

Calculate Work Done in Compression

The work done on a gas during adiabatic compression is given by:\[W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}\].Use the calculated \(V_2\) and known \(V_1\):\[W = \frac{7 \times 0.3486 - 1 \times 1}{1.4 - 1}\]\[W \approx \frac{2.4402 - 1}{0.4} \approx 3.6005\text{ atm-liters}\]. Convert this to Joules (1 atm-liter = 101.3 J):\[W \approx 3.6005 \times 101.3 \approx 364.68\text{ J}\].
04

Calculate Final Temperature After Compression

Use the adiabatic relation for temperature: \[\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}\].Given \(T_1 = 300\text{ K}\), substitute:\[T_2 = 300 \times \left(\frac{1}{0.3486}\right)^{0.4}\]\[T_2 \approx 300 \times 2.8752^{0.4} \approx 300 \times 1.4738 \approx 442.14\text{ K}\].

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Key Concepts

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

Adiabatic Compression
In an adiabatic process, the gas is compressed or expanded without exchanging heat with its surroundings. During adiabatic compression, the gas's volume decreases, while its pressure and temperature increase. This happens without the gas losing or gaining heat. The equation that governs adiabatic processes is \(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\), where \(P\) is pressure, \(V\) is volume, and \(\gamma\) is the adiabatic index.
For diatomic gases like nitrogen and oxygen, \(\gamma\) is typically 1.4. This equation helps us determine the final volume of the gas when the pressure changes. When you pump air into a tire and compress it from 1 atm to 7 atm, the volume significantly decreases. We used this equation to find that the volume after compression is approximately 0.3486 liters, illustrating how much space the compressed air now occupies. Understanding this process is vital in various real-life applications, such as in car engines and refrigeration systems.
Work Done in Gases
Work is done on a gas when its volume changes under pressure, as seen in processes like adiabatic compression. The concept of work done is crucial in thermodynamics and energy transfer. The standard formula for work done during adiabatic compression is \[W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}\]. Here, \(W\) is the work done, and the pressure and volume terms relate to their states before and after compression.
This formula calculates the work based on changes in pressure and volume. In our problem, we found that the work done on the air when compressing it from 1 L to roughly 0.3486 L was about 364.68 Joules. It's essential to know that this energy conversion is why reading the air pressure is crucial when inflating tires, because too much work done can lead to overheating or even bursting of the tire.
Temperature Change in Gases
Temperature change in gases during adiabatic processes is another pivotal concept. When a gas is compressed without heat exchange, its temperature rises. The relationship is expressed as \(\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}\), where \(T_1\) and \(T_2\) are the initial and final temperatures, respectively.
This formula indicates how temperature is highly sensitive to volume changes in an adiabatic process. In our example, with initial air temperature at 300 K, compression increased the temperature to approximately 442.14 K. This rise demonstrates how revving up the conditions without heat loss results in energy being conserved through increased temperature. This principle is also why compressors get hot; the work done on the gas (via pressure and volume changes) raises its temperature, which is notable in many industrial and natural systems.

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Most popular questions from this chapter

A battery is connected in series to a resistor, which is immersed in water (to prepare a nice hot cup of tea). Would you classify the flow of energy from the battery to the resistor as "heat" or "work"? What about the flow of energy from the resistor to the water?

Give an example of a process in which no heat is added to a system, but its temperature increases. Then give an example of the opposite: a process in which heat is added to a system but its temperature does not change.

Give an example to illustrate why you cannot accurately judge the temperature of an object by how hot or cold it feels to the touch.

Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume.

Problem 1.45. As an illustration of why it matters which variables you hold fixed when taking partial derivatives, consider the following mathematical example. Let \(w=x y\) and \(x=y z\) (a) Write \(w\) purely in terms of \(x\) and \(z,\) and then purely in terms of \(y\) and \(z\) (b) Compute the partial derivatives $$ \left(\frac{\partial w}{\partial x}\right)_{y} \quad \text { and } \quad\left(\frac{\partial w}{\partial x}\right)_{z} $$ and show that they are not equal. (Hint: To compute \((\partial w / \partial x)_{y},\) use a formula for \(w\) in terms of \(x\) and \(y,\) not \(z .\) Similarly, compute \((\partial w / \partial x)_{z}\) from a formula for \(w\) in terms of only \(x\) and \(z .\) ) (c) Compute the other four partial derivatives of \(w\) (two each with respect to \(y \text { and } z),\) and show that it matters which variable is held fixed.

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