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Chapter 19: Problem 38

Suppose 1.00 L of a gas with g 1.30, initially at 285 K and 1.00 atm, is suddenly compressed adiabatically to half its initial volume. Find its final (a) pressure and (b) temperature. (c) If the gas is then cooled to 273 K at constant pressure, what is its final volume?

Short Answer

Expert verified
(a) 2.46 atm, (b) 325 K, (c) 0.42 L

Step by step solution

01

Understand the adiabatic process

An adiabatic process is one in which no heat is exchanged with the surroundings. In adiabatic compression, we use the relation \( P_1V_1^\gamma = P_2V_2^\gamma \), where \( \gamma \) is the heat capacity ratio (for air, \( \gamma = 1.30 \)).
02

Use the adiabatic relation for pressure

Starting with the adiabatic equation \( P_1V_1^\gamma = P_2V_2^\gamma \), we need to find \( P_2 \). Given \( P_1 = 1.00 \) atm, \( V_2 = 0.50 \) L, and \( V_1 = 1.00 \) L, compute:\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 1.00 \times \left(\frac{1.00}{0.50}\right)^{1.30} \approx 2.46 \text{ atm} \]
03

Use the adiabatic relation for temperature

We also use the relation \( T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} \) to find \( T_2 \). Given \( T_1 = 285 \) K:\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 285 \times \left(\frac{1.00}{0.50}\right)^{0.30} \approx 325 \text{ K} \]
04

Cooling at constant pressure

At constant pressure, we use Charles's Law \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). We want to find \( V_2 \) when \( T_2 = 273 \) K, maintaining pressure at the final adiabatically compressed pressure:\[ V_2 = V_1 \times \frac{T_2}{T_1} = 0.50 \times \frac{273}{325} \approx 0.42 \text{ L} \]

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Key Concepts

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

Heat Capacity Ratio
The heat capacity ratio, often symbolized by \( \gamma \), is a vital concept in thermodynamics, especially when dealing with gases. It is defined as the ratio of two specific heats: the specific heat at constant pressure \( C_p \) and the specific heat at constant volume \( C_v \). The formula is \( \gamma = \frac{C_p}{C_v} \).
  • This ratio is crucial because it determines how a gas behaves under adiabatic processes鈥攚here no heat is transferred into or out of the system.
  • A higher heat capacity ratio generally means the gas will have a greater change in temperature and pressure during compression or expansion.
  • For monoatomic gases like helium, \( \gamma \) is typically higher (around 1.67), while for diatomic gases such as oxygen, it's lower (around 1.4).
In the problem above, the heat capacity ratio used was 1.30, confirming it was likely a diatomic gas, like air, being compressed adiabatically.
Adiabatic Compression
Adiabatic compression is a process where a gas is compressed in such a way that no heat is exchanged with its surroundings. In this process, all the work done on the gas results in an increase in its internal energy, often increasing the temperature and pressure of the gas.
  • During adiabatic compression, according to the law \( P_1V_1^\gamma = P_2V_2^\gamma \), pressure and volume are related uniquely.
  • The formula shows how they adjust depending on their initial and final states.
  • In our example, the initial pressure and volume were 1.00 atm and 1.00 L, respectively, and they shifted to 2.46 atm when the volume was compressed to 0.50 L.
  • The absence of heat exchange means that temperature changes occur due to changes in internal energy, as seen by the increase to 325 K.
This example showcases how using the heat capacity ratio can predict the final states within adiabatic processes.
Charles's Law
Charles's Law is an important principle in thermodynamics, especially when considering the behavior of gases at constant pressure. This law states that the volume of a gas is directly proportional to its temperature (in Kelvin), provided the pressure is constant.
  • The mathematical expression is \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
  • It describes how gases expand when heated and contract when cooled at a constant pressure.
In the context of our problem, after adiabatically compressing the gas, it was cooled to 273 K at the final adiabatic pressure. The change in temperature at constant pressure allowed us to find the new volume of the gas, which was calculated to be approximately 0.42 L. Charles's Law ensures you're able to predict how a gas will occupy space when subjected to temperature changes.
Gas Laws
Gas laws form the backbone of our understanding of how gases behave under varying conditions of pressure, volume, and temperature. Key principles include Boyle's Law, Charles's Law, and the Ideal Gas Law, often formulated as \( PV = nRT \), where \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.
  • These laws help describe the relationships between different variables affecting gases.
  • The adiabatic process mentioned in the example above is another crucial element, not explicitly part of these basic laws but extending from them through the specific heat ratio.
  • Understanding gas laws equips you to calculate changes in state variables due to processes like compression or heating.
In practical terms, these laws allow the transition from theoretical equations to real-world applications, making sense of pressure cookers, internal combustion engines, and even weather patterns.

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Most popular questions from this chapter

Two containers are at the same temperature. The first contains gas with pressure p1, molecular mass m1, and rms speed vrms1. The second contains gas with pressure 1.5p1, molecular mass m2, and average speed vavg2 2.0vrms1. Find the mass ratio m1/m2.

The temperature of 4.50 mol of an ideal monatomic gas is raised 27.0 K at constant volume.What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change Eint in the internal energy of the gas, and (d) the change K in the average kinetic energy per atom?

Opening champagne. In a bottle of champagne, the pocket of gas (primarily carbon dioxide) between the liquid and the cork is at pressure of \(p_{i}=4.00 \mathrm{~atm}\). When the cork is pulled from the bottle, the gas undergoes an adiabatic expansion until its pressure matches the ambient air pressure of \(1.00\) atm. Assume that the ratio of the molar specific heats is \(y=\frac{4}{3}\). If the gas has initial temperature \(T_{i}=5.00^{\circ} \mathrm{C}\), what is its temperature at the end of the adiabatic expansion?

A beam of hydrogen molecules (H2) is directed toward a wall, at an angle of 32 with the normal to the wall. Each molecule in the beam has a speed of 1.0 km/s and a mass of 3.3 1024 g. The beam strikes the wall over an area of 2.0 cm2 , at the rate of 4.0 1023 molecules per second. What is the beam鈥檚 pressure on the wall?

Suppose 1.80 mol of an ideal gas is taken from a volume of 3.00 m3 to a volume of 5.50 m3 via an isothermal expansion at 30C. (a) How much energy is transferred as heat during the process, and (b) is the transfer to or from the gas?
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