Chapter 19: Problem 10
Five moles of an ideal monatomic gas with an initial temperature of 127\(^\circ\)C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?
Short Answer
Expert verified
The final temperature of the gas is 390.5 K.
Step by step solution
01
Convert Initial Temperature to Kelvin
The initial temperature of the gas is given as 127\(^\circ\)C. To convert this to Kelvin, we add 273.15. Thus, \(T_i = 127 + 273.15 = 400.15\,\text{K}\).
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 absorbed, and \( W \) is the work done by the system. Here, \( Q = 1500\, J \) and \( W = 2100\, J \). Thus, \( \Delta U = 1500 - 2100 = -600\, J \).
03
Relate Internal Energy Change to Temperature Change
For an ideal monatomic gas, the change in internal energy can also be expressed using \( \Delta U = \frac{3}{2}nR\Delta T \), where \( n \) is the number of moles and \( R \) is the ideal gas constant \( 8.314\, J/(mol\cdot K) \).
04
Solve for Change in Temperature \( \Delta T \)
Using \( \Delta U = \frac{3}{2}nR\Delta T \), we have: \[ -600 = \frac{3}{2} \times 5 \times 8.314 \times \Delta T \]Solve for \( \Delta T \):\[ \Delta T = \frac{-600}{\frac{3}{2} \times 5 \times 8.314} \]\[ \Delta T \approx -9.65\, K \]
05
Calculate the Final Temperature
Given \( T_i = 400.15\, K \) and \( \Delta T = -9.65\, K \), the final temperature \( T_f = T_i + \Delta T = 400.15 - 9.65 = 390.5\, K \).
06
Convert Final Temperature to Celsius (if needed)
To convert the final temperature from Kelvin to Celsius: \[ T_f(\degree C) = 390.5 - 273.15 = 117.35\, \degree C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
first law of thermodynamics
The first law of thermodynamics is a fundamental principle that helps us understand energy changes within a system. It's closely related to the conservation of energy. The law states: \( \Delta U = Q - W \), where:
- \( \Delta U \) is the change in internal energy of the system.
- \( Q \) is the heat transferred into the system (positive if absorbed).
- \( W \) is the work done by the system on its surroundings (positive if the system does work).
internal energy
Internal energy is associated with the microscopic components of a system, like molecules and atoms in a gas. For an ideal monatomic gas, internal energy depends solely on its temperature and is given by: \( U = \frac{3}{2} nRT \).
This equation links internal energy directly to the amount of heat added and the work performed.
This equation links internal energy directly to the amount of heat added and the work performed.
- \( n \) is the number of moles.
- \( R \) is the ideal gas constant, \( 8.314\, J/(mol\cdot K) \).
- \( T \) is the absolute temperature in Kelvin.
temperature conversion
Converting temperature from Celsius (°C) to Kelvin (K) is often necessary in thermodynamics. This conversion ensures that calculations align with related formulas that require absolute temperatures in Kelvin.
The conversion formula is straightforward: \( T(K) = T(°C) + 273.15 \).
For this exercise:
The conversion formula is straightforward: \( T(K) = T(°C) + 273.15 \).
For this exercise:
- The initial temperature is 127°C. Add 273.15 to obtain 400.15 K.
- After calculations, the final temperature of 390.5 K can be converted back to Celsius for practical understanding if needed, using \( T(°C) = T(K) - 273.15 \).
- This results in a final temperature of 117.35°C.
monatomic gas
Monatomic gases consist of individual atoms, like helium or neon, and have properties that simplify thermodynamics calculations. For a monatomic ideal gas, the internal energy and temperature are directly related.
- The equation \( U = \frac{3}{2}nRT \) highlights this relationship.
- These gases follow simple behavior due to their lack of molecular bonds, unlike diatomic or polyatomic gases.