Chapter 18: Problem 53
External forces compress 21 mol of ideal monatomic gas. During the process, the gas transfers \(15 \mathrm{kJ}\) of heat to its surroundings, yet its temperature rises by \(160 \mathrm{K}\). How much work was done on the gas?
Short Answer
Expert verified
The work done on the gas is 81.59 kJ.
Step by step solution
01
– Calculate the change in internal energy
Firstly we calculate how much energy has changed in the system. For an ideal monatomic gas, the change in internal energy \( \Delta U \) can be calculated by the formula \( \Delta U = \frac{3}{2} Nk\Delta T \) where N is the number of moles, k is the Boltzmann constant and \( \Delta T \) is the change in temperature. Plugging in the given numbers, \( \Delta U = \frac{3}{2} \cdot 21 \cdot 8.3145 \cdot 160 = 66.5916 \, kJ \).
02
– Calculate the work done on the gas
To find out the work done on the gas, we re-arrange the first law of thermodynamics to W = \( \Delta U - Q \). It's important to note that the given heat Q is in reverse, it's released to the surroundings, hence it's negative (-15 kJ). So the formula becomes W = 66.5916 - (-15) = 81.5916 kJ.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Internal Energy Change
Understanding the concept of internal energy change is pivotal when studying thermodynamics. Internal energy, represented by the symbol \( U \), is the sum of potential and kinetic energies of all the particles within a system. This energy can be altered by heating or doing work on the system. Specifically, in the context of an ideal monatomic gas, we're referring to the kinetic energy of the atoms as they move and collide with each other.
In thermodynamics, the change in internal energy \( \Delta U \) occurs when there is an input or release of heat, or when work is done on or by the system. The exercise provided explores this by compressing a monatomic gas, which increases its internal energy. For an ideal monatomic gas, this change is given by the formula \( \Delta U = \frac{3}{2} Nk\Delta T \), where \( N \) stands for the number of moles, \( k \) is the Boltzmann constant, and \( \Delta T \) signifies the change in temperature.
In our specific example, compressing the gas increases its temperature, demonstrating that work is done on the system, effectively increasing its internal energy. The equation provided gives us a quantitative interpretation of this energy change.
In thermodynamics, the change in internal energy \( \Delta U \) occurs when there is an input or release of heat, or when work is done on or by the system. The exercise provided explores this by compressing a monatomic gas, which increases its internal energy. For an ideal monatomic gas, this change is given by the formula \( \Delta U = \frac{3}{2} Nk\Delta T \), where \( N \) stands for the number of moles, \( k \) is the Boltzmann constant, and \( \Delta T \) signifies the change in temperature.
In our specific example, compressing the gas increases its temperature, demonstrating that work is done on the system, effectively increasing its internal energy. The equation provided gives us a quantitative interpretation of this energy change.
Ideal Monatomic Gas
An ideal monatomic gas is a simplified model used in physics and chemistry to describe the behavior of gases. In this model, the gas consists of single atoms (hence, 'monatomic'), and several assumptions are made for simplification:
This concept is crucial to solving many thermodynamic problems as it allows predictions and calculations under idealized conditions, which can then be adjusted or accounted for with more complex real-world behavior. Our exercise captures this through the quantitative calculation of the change in internal energy due to a temperature rise during compression.
- The atoms are considered as point particles with mass but no volume.
- There are no intermolecular forces except during elastic collisions.
- The collisions between atoms are perfectly elastic, meaning there is no loss of kinetic energy.
- The motion of the atoms follows the laws of classical mechanics.
This concept is crucial to solving many thermodynamic problems as it allows predictions and calculations under idealized conditions, which can then be adjusted or accounted for with more complex real-world behavior. Our exercise captures this through the quantitative calculation of the change in internal energy due to a temperature rise during compression.
Work Done on Gas
The term 'work done on gas' in thermodynamics refers to the energy transferred to a gas when an external force acts upon it, typically during compression. Work is a form of energy transfer and can be either done by the system (the gas) when it expands, or on the system when it is compressed.
In the context of the first law of thermodynamics, the work done on the gas is related to the change in internal energy and the exchange of heat. The law states that the change in internal energy of a system is equal to the sum of the heat added to the system and the work done on the system: \( \Delta U = Q + W \). To find the work done on the gas when heat is exchanged, we manipulate the formula to \( W = \Delta U - Q \).
In our specific problem, the external forces doing the work result in the compression of the gas, which is denoted by a positive value when calculating work done on the system. As the exercise reveals, with the provided heat transfer and temperature change, using the first law gives us a clear value for the work done on the gas, contributing to our overall understanding of the energy dynamics within the system.
In the context of the first law of thermodynamics, the work done on the gas is related to the change in internal energy and the exchange of heat. The law states that the change in internal energy of a system is equal to the sum of the heat added to the system and the work done on the system: \( \Delta U = Q + W \). To find the work done on the gas when heat is exchanged, we manipulate the formula to \( W = \Delta U - Q \).
In our specific problem, the external forces doing the work result in the compression of the gas, which is denoted by a positive value when calculating work done on the system. As the exercise reveals, with the provided heat transfer and temperature change, using the first law gives us a clear value for the work done on the gas, contributing to our overall understanding of the energy dynamics within the system.