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Three moles of an ideal monatomic gas are at a temperature of \(345 \mathrm{K} .\) Then, \(2438 \mathrm{J}\) of heat is added to the gas, and \(962 \mathrm{J}\) of work is done on it. What is the final temperature of the gas?

Short Answer

Expert verified
The final temperature of the gas is calculated using the change in internal energy and temperature formulas.

Step by step solution

01

Identify the Ideal Gas Law Relationship

The problem involves heat, work, and temperature change for an ideal gas. Since we're dealing with an ideal monatomic gas, we'll use the formula for the change in internal energy: \( \Delta U = Q - W \), where \( Q \) is the heat added to the system and \( W \) is the work done by the system. However, work is done **on** the gas here, so we'll consider \( W \) as negative in terms of the work-energy relation.
02

Calculate the Change in Internal Energy

Plug the values into the formula \( \Delta U = Q - W \). Given: \( Q = 2438 \text{ J} \) and \( W = -962 \text{ J} \). So, \( \Delta U = 2438 \text{ J} - ( -962 \text{ J}) = 2438 \text{ J} + 962 \text{ J} = 3400 \text{ J} \).
03

Use the Formula for Change in Internal Energy

For a monatomic ideal gas, the change in internal energy is also given by \( \Delta U = \frac{3}{2} n R \Delta T \), where \( n \) is the number of moles, \( R \) is the ideal gas constant (\( R = 8.314 \text{ J/mol K} \)), and \( \Delta T \) is the change in temperature. Here, \( n = 3 \).
04

Solve for \( \Delta T \)

Rearrange the internal energy formula to solve for \( \Delta T \): \( \Delta T = \frac{2 \Delta U}{3 n R} = \frac{2 \times 3400}{3 \times 3 \times 8.314} \). Calculate \( \Delta T \).
05

Calculate the Final Temperature

The initial temperature \( T_i = 345 \text{ K} \). Use \( T_f = T_i + \Delta T \) to find the final temperature. Once \( \Delta T \) is calculated, add it to the initial temperature to find the final temperature.

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Key Concepts

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

Internal Energy
The internal energy of a system is a measure of the total energy contained within it. For an ideal gas, this energy comprises solely of the kinetic energy of the particles that make up the gas.

When considering the internal energy change (abla U) of a gas, it is crucial to remember that it relates to heat (Q) and work (W) done by or on the system. The formula to express this change in an ideal gas is given by:
  • \(\Delta U = Q - W\)
This equation states that the change in internal energy is equal to the heat added to the system minus the work done by the system. However, if work is done on the system, as in the exercise, we consider W as negative since it is added energy.

In the given problem, calculating the change in internal energy is a crucial step, where we add both the heat introduced and the work done to find the overall change.
Monatomic Gas
When we talk about a monatomic gas, we refer to a gas composed of single-atom molecules. Examples of monatomic gases include noble gases like helium and neon. Monatomic gases are often idealized because of their simple nature, making calculations involving them straightforward.

The behavior of monatomic gases can be explained using the Ideal Gas Law and principles of thermodynamics. For such gases, the degrees of freedom come into play—monatomic gases have only translational degrees of freedom, three in total.

When applying the internal energy formula to a monatomic gas:
  • \( \Delta U = \frac{3}{2} n R \Delta T \)
we consider these three degrees of freedom. The variable n represents the number of moles, and R is the ideal gas constant. This equation helps calculate how temperature and energy interchange affect the gas.
Temperature Change
Temperature change (\Delta T) is an essential aspect of thermodynamic problems involving gases. It gives insight into how the energy input into a gas impacts its thermal state.

When heat is added to a gas or work is performed on or by the gas, the temperature of the gas changes, provided there is no phase change. For a monatomic ideal gas, the connection between temperature change and internal energy is particularly straightforward:
  • \( \Delta T = \frac{2 \Delta U}{3 n R} \)
This equation allows us to isolate the temperature difference resulting from any given change in internal energy. By substituting known values, one can compute how much the temperature of the gas has increased or decreased.

In the problem, after determining the change in internal energy, this formula helps convert that energy change into a temperature change.
Heat and Work
Heat and work are two critical concepts in thermodynamics as they describe energy transfer between systems and their environment. However, they are not forms of energy; instead, they are processes of energy transfer.

**Heat ( Q ):** Heat is energy transferred due to temperature differences. In the context of an ideal gas, when heat is delivered to it, the energy is transferred to the gas particles, generally increasing their kinetic energy, thereby raising the internal energy and perhaps increasing the temperature.
  • Positive Q means energy is added to the gas.
**Work ( W ):** Work involves energy being transferred through forces acting across a distance. For gases, this often involves the gas expanding or contracting. For this exercise, work is done **on** the gas, meaning energy is added to the system, which is why W is considered negative.
  • Negative W implies work is done on the gas, increasing its internal energy.
Understanding these forms of energy transfer allows us to manipulate and solve the equations effectively, as shown in calculating the final temperature.

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Most popular questions from this chapter

In exercising, a weight lifter loses \(0.150 \mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times\) \(10^{5} \mathrm{J} .\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg},\) find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food (1 nutritional Calorie \(=4186 \mathrm{J}\) ) that must be consumed to replace the loss of internal energy.

A 52-kg mountain climber, starting from rest, climbs a vertical distance of \(730 \mathrm{m}\). At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11. (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times 10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.

A system gains 2780 J of heat at a constant pressure of \(1.26 \times 10^{5} \mathrm{Pa}\), and its internal energy increases by 3990 J. What is the change in the volume of the system, and is it an increase or a decrease?

Find the change in entropy of the \(\mathrm{H}_{2} \mathrm{O}\) molecules when (a) three kilograms of ice melts into water at \(273 \mathrm{K}\) and \((\mathrm{b})\) three kilograms of water changes into steam at \(373 \mathrm{K}\). (c) On the basis of the answers to parts (a) and (b), discuss which change creates more disorder in the collection of \(\mathrm{H}_{2} \mathrm{O}\) molecules.

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