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Chapter 15: Problem 37

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by \(75 \mathrm{K}\). In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

Short Answer

Expert verified
The temperature of the second sample increases by 45 K.

Step by step solution

01

Understand Heat Capacities for Monatomic Ideal Gas

For a monatomic ideal gas, the molar heat capacity at constant volume, \( C_V \), is \( \frac{3}{2}R \), and at constant pressure, \( C_P \), is \( \frac{5}{2}R \), where \( R \) is the gas constant. These capacities tell us how much heat is required to change the temperature of the gas by one Kelvin.
02

Recall Heat Addition Equation at Constant Volume

The heat added at constant volume is related to the temperature change by the formula \( Q = n C_V \Delta T_{V} \). Here, \( n \) is the amount of substance, \( C_V \) is the heat capacity at constant volume, and \( \Delta T_{V} \) is the temperature change. Given \( \Delta T_{V} = 75 \) K.
03

Set Up Equation for Second Sample at Constant Pressure

Similarly, for the second sample, heat is added at constant pressure, and the relationship is \( Q = n C_P \Delta T_{P} \). We know \( Q \) is the same as in the first case, and we need to find \( \Delta T_{P} \).
04

Compare Two Scenarios

Since the same heat \( Q \) is added to both samples, equate the equations: \( n C_V \Delta T_{V} = n C_P \Delta T_{P} \). Simplify to find \( \Delta T_{P} \).
05

Solve for Temperature Change at Constant Pressure

Substitute \( C_V = \frac{3}{2}R \) and \( C_P = \frac{5}{2}R \) into the equality derived: \( \frac{3}{2}R \times 75 = \frac{5}{2}R \times \Delta T_{P} \). Solving gives \( \Delta T_{P} = \frac{3}{5} \times 75 = 45 \) K.

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Key Concepts

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

Heat Capacity
Heat capacity is a measure of the amount of heat needed to change the temperature of a substance. In thermodynamics, heat capacity is crucial for understanding how substances react to added heat.
Heat capacity can differ based on how the heat is added to the system. For instance, it's different when the gas is held at constant volume versus constant pressure:
  • Constant Volume: The heat capacity at constant volume, denoted by \( C_V \), represents how much heat is required to raise the temperature of the gas without changing its volume. For a monatomic ideal gas, \( C_V = \frac{3}{2}R \), where \( R \) is the universal gas constant.
  • Constant Pressure: The heat capacity at constant pressure, \( C_P \), measures the heat required at a constant pressure. It is larger than \( C_V \) because the gas can expand, doing work on its surroundings. For a monatomic ideal gas, \( C_P = \frac{5}{2}R \).
Monatomic Ideal Gas
A monatomic ideal gas is a theoretical model for gases composed of single-atom molecules that perfectly follow the assumptions of the ideal gas law. In reality, gases may deviate from these assumptions under high pressure or low temperature, but this model provides a useful approximation for many problems.
Monatomic gases, like helium or neon, possess certain characteristics making calculations easier:
  • The particles do not interact with each other, except for elastic collisions.
  • They possess translational kinetic energy, and their rotational and vibrational energies are negligible.
  • The heat capacities \( C_V \) and \( C_P \) are derived from these assumptions, leading to specific values for calculations.
Temperature Change
The temperature change in a gas depends on how the heat is added to the system and the specific heat capacity of the gas. For a monatomic ideal gas, adding heat at constant volume or constant pressure will yield different results.
When energy in the form of heat \( Q \) is added to a gas:
  • At constant volume, the entire energy goes into increasing the temperature, without causing any work to be done by the gas. The formula \( Q = n C_V \Delta T_{V} \) shows us that the temperature change \( \Delta T_{V} = 75 \) K tells us about the heat capacity \( C_V \).
  • At constant pressure, some energy goes into doing work as the gas expands, and the temperature change is smaller. Hence, the formula \( Q = n C_P \Delta T_{P} \) helps us calculate when the pressure remains constant. The example from earlier gives \( \Delta T_{P} = 45 \) K, indicating a different impact due to \( C_P \).
Constant Volume and Pressure Processes
Processes involving constant volume and constant pressure are common in thermodynamics and help illustrate how gases behave under different conditions. Understanding these processes aids in solving many physics and chemistry problems.
  • Constant Volume Process: Here, the volume doesn't change despite heat being added or removed. This means \( \Delta V = 0 \), and so no work \( W \) is done (since work \( W = P \Delta V \)). In practical terms, the system heats up or cools down purely because its internal energy is changing.
  • Constant Pressure Process: In this case, the pressure remains unchanged, but the gas can expand or contract. The work done by the gas (or on the gas) is not zero, as it may increase volume (\( W = P \Delta V \) is not zero). Thus, internal energy changes are partly translated into work, causing a smaller temperature change for the same heat input compared to constant volume. The example, where temperature change is only 45 K here, illustrates such behavior efficiently.

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Most popular questions from this chapter

A monatomic ideal gas has an initial temperature of \(405 \mathrm{K}\). This gas expands and does the same amount of work whether the expansion is adiabatic or isothermal. When the expansion is adiabatic, the final temperature of the gas is \(245 \mathrm{K}\). What is the ratio of the final to the initial volume when the expansion is isothermal?

A system does \(4.8 \times 10^{4} \mathrm{J}\) of work, and \(7.6 \times 10^{4} \mathrm{J}\) of heat flows into the system during the process. Find the change in the internal energy of the system.

An engine has a hot-reservoir temperature of \(950 \mathrm{K}\) and a cold- reservoir temperature of \(620 \mathrm{K}\). The engine operates at three-fifths maximum efficiency. What is the efficiency of the engine?

When a .22-caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure cause the force that pushes the bullet through the barrel. The barrel has length of \(0.61 \mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} .\) A bulle (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of \(370 \mathrm{m} / \mathrm{s}\) after passing through th barrel. Ignore friction and determine the average pressure of the expanc ing gas.

Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 200 students. Assume that the metabolic rate of generating heat is \(130 \mathrm{W}\) for each student and that the heat accumulates during a fifty-minute lecture. In addition, assume that the air has a molar specific heat of \(C_{V}=\frac{5}{2} R\) and that the room (volume \(=1200 \mathrm{m}^{3},\) initial pressure \(=1.01 \times 10^{5} \mathrm{Pa}\), and initial temperature \(=21^{\circ} \mathrm{C}\) ) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise during a lecture?
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