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Magazine Chapter 15: Problem 22
Three moles of neon expand isothermally to 0.250 from \(0.100 \mathrm{m}^{3}\). Into the gas flows \(4.75 \times 10^{3} \mathrm{J}\) of heat. Assuming that neon is an ideal gas, find its temperature.
Short Answer
Expert verified
The temperature of the gas is approximately 206.8 K.
Step by step solution
01
Understand the Isothermal Process
In an isothermal process, the temperature of the system remains constant. For an ideal gas, this means that the internal energy change \((\Delta U)\) is zero since internal energy is only a function of temperature.
02
Apply the First Law of Thermodynamics
The first law of thermodynamics is given by \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. Since \( \Delta U = 0 \) for an isothermal process, it follows that \( Q = W \). The heat added \( Q = 4.75 \times 10^3 \mathrm{J} \) is equal to the work done by the gas.
03
Use the Work Formula for Isothermal Expansion
For an ideal gas undergoing isothermal expansion, the work done \( W \) is given by \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \) where \( n = 3 \) moles, \( R = 8.314 \mathrm{J\,mol}^{-1}\,\mathrm{K}^{-1} \), \( V_f = 0.250 \mathrm{m}^3 \), and \( V_i = 0.100 \mathrm{m}^3 \).
04
Solve for Temperature \( T \)
Using the equation \( Q = nRT \ln\left(\frac{V_f}{V_i}\right) \), we substitute the known values: \[ 4.75 \times 10^3 = 3 \times 8.314 \times T \times \ln(\frac{0.250}{0.100}) \]Calculate \( \ln(\frac{0.250}{0.100}) = \ln(2.5) \approx 0.916 \). Thus, \[ 4.75 \times 10^3 = 3 \times 8.314 \times T \times 0.916 \]This simplifies to \[ 4.75 \times 10^3 = 22.97 \times T \] Finally, \[ T = \frac{4.75 \times 10^3}{22.97} \approx 206.8 \mathrm{K} \]
05
Conclusion
The temperature of the ideal gas, neon, under isothermal expansion, given the heat flow into the system is approximately \( 206.8 \mathrm{K} \).
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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 simplifies the behavior of gases. Defined by the ideal gas law, PV = nRT, where
This equation helps us predict how gases behave under different conditions of pressure, volume, and temperature. The concept assumes no intermolecular forces between gas molecules and that the volume of the molecules is negligible compared to the container.
This model works well at high temperatures and low pressures, where real gases behave similarly to ideal ones. In our exercise, neon is treated as an ideal gas, which allows us to use formulas and principles derived from the ideal gas law without additional complexities.
- P is the pressure,
- V is the volume,
- n is the number of moles,
- R is the universal gas constant, and
- T is the temperature in Kelvin.
This equation helps us predict how gases behave under different conditions of pressure, volume, and temperature. The concept assumes no intermolecular forces between gas molecules and that the volume of the molecules is negligible compared to the container.
This model works well at high temperatures and low pressures, where real gases behave similarly to ideal ones. In our exercise, neon is treated as an ideal gas, which allows us to use formulas and principles derived from the ideal gas law without additional complexities.
First Law of Thermodynamics
The first law of thermodynamics is a fundamental principle that describes the conservation of energy in a system. It states:
For isothermal processes involving an ideal gas, the temperature鈥攁nd thus the internal energy鈥攔emains constant. This implies \( \Delta U = 0 \), simplifying the equation to \( Q = W \).
Understanding this relationship is crucial for solving problems involving energy transfers where temperature is constant, as it allows us to focus on the balance between heat energy and work.
- \( \Delta U = Q - W \)
- \( \Delta U \) is the change in internal energy,
- \( Q \) is the heat added to the system, and
- \( W \) is the work done by the system.
For isothermal processes involving an ideal gas, the temperature鈥攁nd thus the internal energy鈥攔emains constant. This implies \( \Delta U = 0 \), simplifying the equation to \( Q = W \).
Understanding this relationship is crucial for solving problems involving energy transfers where temperature is constant, as it allows us to focus on the balance between heat energy and work.
Work Done in Isothermal Expansion
In an isothermal expansion, an ideal gas does work as it expands at constant temperature. The work done by the gas is expressed as:
This formula shows that the work done depends on the amount of gas ( n ), the temperature, and the change in volume. Since \( W = Q \) in an isothermal process for an ideal gas, it illustrates that all the heat added to the system goes into doing work on its surroundings without changing internal energy.
In our example, we used known values to find the temperature, illustrating how these principles are applied in practice.
- \( W = nRT \ln\left( \frac{V_f}{V_i} \right) \)
- \( n \) is the number of moles,
- \( R \) is the gas constant,
- \( T \) is the temperature,
- \( V_f \) is the final volume,
- \( V_i \) is the initial volume.
This formula shows that the work done depends on the amount of gas ( n ), the temperature, and the change in volume. Since \( W = Q \) in an isothermal process for an ideal gas, it illustrates that all the heat added to the system goes into doing work on its surroundings without changing internal energy.
In our example, we used known values to find the temperature, illustrating how these principles are applied in practice.
Internal Energy Change
Internal energy refers to the total energy stored within a system due to molecular motion and interactions. For an ideal gas, the internal energy depends solely on temperature, not on pressure or volume.
This concept is crucial in understanding why the first law of thermodynamics simplifies so nicely for isothermal processes of ideal gases.
Instead of worrying about complex energy exchanges, we know that any heat added to or removed from the system manifests as work done, not as a change in internal energy. In practical terms, this is why calculating temperature in processes like the one described becomes straightforward using the given relationships.
- In isothermal processes, the temperature does not change.
- This means there is no change in internal energy (\( \Delta U = 0 \)).
This concept is crucial in understanding why the first law of thermodynamics simplifies so nicely for isothermal processes of ideal gases.
Instead of worrying about complex energy exchanges, we know that any heat added to or removed from the system manifests as work done, not as a change in internal energy. In practical terms, this is why calculating temperature in processes like the one described becomes straightforward using the given relationships.
