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The temperature of \(2.00 \mathrm{~mol}\) of an ideal monatomic gas is raised \(15.0 \mathrm{~K}\) at constant volume. What are (a) the work \(W\) done by the gas, (b) the energy transferred as heat \(Q,\) (c) the change \(\Delta E_{\text {int }}\) in the internal energy of the gas, and (d) the change \(\Delta K\) in the average kinetic energy per atom?

Short Answer

Expert verified
(a) \( W = 0 \text{ J} \); (b) \( Q = 374.13 \text{ J} \); (c) \( \Delta E_{\text{int}} = 374.13 \text{ J} \); (d) \( \Delta K = 3.105 \times 10^{-22} \text{ J} \).

Step by step solution

01

Understanding the Problem

We have an ideal monatomic gas with a temperature increase of \(15.0 \text{ K}\) at constant volume. We need to find the work done by the gas, the energy transfer as heat, the change in internal energy, and the change in average kinetic energy per atom.
02

Calculating Work Done, \(W\)

At constant volume, the work done by the gas on its surroundings is zero because work done by a gas is defined as \( W = P \Delta V \). Since \( \Delta V = 0 \), \( W = 0 \text{ J} \).
03

Using First Law of Thermodynamics for Heat \( Q \)

The first law of thermodynamics states \( \Delta E_{\text{int}} = Q - W \). As \( W = 0 \) in this case, \( Q = \Delta E_{\text{int}} \). Where \( \Delta E_{\text{int}} \) can be calculated using \( \Delta E_{\text{int}} = \frac{3}{2} n R \Delta T \).
04

Calculating Change in Internal Energy \( \Delta E_{\text{int}} \)

For a monatomic ideal gas, \( \Delta E_{\text{int}} \) is given by \( \Delta E_{\text{int}} = \frac{3}{2} n R \Delta T \). Substitute \( n = 2.00 \text{ mol} \), \( \Delta T = 15.0 \text{ K} \), and \( R = 8.314 \text{ J/mol}\cdot\text{K} \) to get \( \Delta E_{\text{int}} = \frac{3}{2} \times 2.00 \times 8.314 \times 15.0 = 374.13 \text{ J} \).
05

Calculating Energy Transferred as Heat \( Q \)

Since \( Q = \Delta E_{\text{int}} \), we have \( Q = 374.13 \text{ J} \).
06

Determining Change in Average Kinetic Energy \( \Delta K \)

The change in the average kinetic energy per atom of an ideal gas is given by \( \Delta K = \frac{3}{2} k_B \Delta T \), where \( k_B \) (Boltzmann constant) is \( 1.38 \times 10^{-23} \text{ J/K} \). For \( \Delta T = 15.0 \text{ K} \), \( \Delta K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 15.0 = 3.105 \times 10^{-22} \text{ J} \).

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical model that helps us understand how gases behave under various conditions. The concept of an ideal gas simplifies real gas behavior by assuming that gas particles have:
  • No volume of their own. They are point particles.
  • No inter-particle forces, meaning they do not attract or repel each other.
  • Completely elastic collisions, so no energy is lost when particles collide.
With these assumptions, calculations involving gases become more straightforward. We use the ideal gas law as a key equation for these calculations:\[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. In the context of the exercise above, the gas remains at constant volume, so no work is done by the gas, simplifying the calculations considerably.
Internal Energy
The internal energy of a system refers to the total energy contained within it. For an ideal gas, the internal energy is directly related to its temperature. The higher the temperature, the greater the internal energy because the gas particles move faster.
For a monatomic ideal gas, the change in internal energy \( \Delta E_{\text{int}} \) is calculated using:\[\Delta E_{\text{int}} = \frac{3}{2} n R \Delta T\]where \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( \Delta T \) is the change in temperature. This formula illustrates that the internal energy depends on temperature change, rather than on pressure or volume changes, under constant volume conditions.
In the exercise mentioned, this formula was used to find the change in internal energy, reinforcing the foundational understanding of temperature’s role in internal energy for an ideal gas.
Work in Thermodynamics
In thermodynamics, work is the energy transferred when a force is applied to move something. For gases, work is often associated with volume changes. When a gas expands or contracts, it does work on its surroundings or work is done on it.
The expression for work done by a gas is given by:\[ W = P \Delta V \]where \( W \) is work, \( P \) is pressure, and \( \Delta V \) is the change in volume. However, in cases where the volume remains constant, as in our exercise, \( \Delta V = 0 \) which means \( W = 0 \). Thus, no work is done by the gas or on the gas. This concept simplifies the analysis as understanding work in thermodynamics often revolves around changes in state variables such as volume, pressure, and temperature.
Kinetic Theory of Gases
The kinetic theory of gases provides insights into the behavior of gas particles and underpins many principles in thermodynamics. It views gas particles as in constant, random motion, with their speed and kinetic energy related to the temperature of the gas.
Key postulates include:
  • All gas particles are in continuous, random motion.
  • Collisions between gas particles are perfectly elastic, meaning no energy is lost.
  • The average kinetic energy of gas particles is directly proportional to the absolute temperature.
In the example exercise, the average kinetic energy change per atom \( \Delta K \) was calculated using:\[\Delta K = \frac{3}{2} k_B \Delta T\]where \( k_B \) is the Boltzmann constant and \( \Delta T \) is the change in temperature. This shows that temperature increases directly lead to increases in the kinetic energy of gas particles, aligning with real-world observations that hotter gases have particles moving more vigorously.

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