Problem 36 - An Introduction to Thermal Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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 \( \gamma = \frac{7}{5} = 1.4\).

Step 2: 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 \(V_2\), \[V_2 = \left(\frac{1}{7}\right)^{\frac{1}{1.4}}\] \[V_2 \approx 0.3486\text{ liters}\].

Step 3: 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):\[W \approx 3.6005 \times 101.3 \approx 364.68\text{ J}\].

Step 4: 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:\[T_2 = 300 \times \left(\frac{1}{0.3486}\right)^{0.4}\]\[T_2 \approx 300 \times 2.8752^{0.4} \approx 300 \times 1.4738 \approx 442.14\text{ K}\].

Key Concepts

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.