Problem 20
Question
One mole of a non-ideal gas undergoes a change of state ( \(2.0\) atm, 3.0L, 95K) \(\rightarrow(4.0\) atm \(, 5.0 \mathrm{~L}, 245 \mathrm{~K})\) with a change in internal energy, \(\Delta U=30.0 \mathrm{~L} \mathrm{~atm} .\) The change in enthalpy \((\Delta H)\) of the process in \(\mathrm{L}\) atm is [2002S] (a) \(40.0\) (b) \(42.3\) (c) \(44.0\) (d) not defined, because pressure is not constant
Step-by-Step Solution
Verified Answer
The change in enthalpy \( \Delta H \) is 44.0 L atm, corresponding to option (c).
1Step 1: Understanding the Relationship Between Enthalpy and Internal Energy
The change in enthalpy (\( \Delta H \)) for a process can be calculated using the relationship: \[ \Delta H = \Delta U + \Delta (PV) \]where \( \Delta U \) is the change in internal energy and \( \Delta (PV) \) is the change in the product of pressure and volume. Here, both initial and final states are provided to calculate these terms.
2Step 2: Calculating Initial and Final PV Values
The initial state has pressure \( P_1 = 2.0 \) atm and volume \( V_1 = 3.0 \) L, so the initial \( PV \) value is:\[ (PV)_1 = P_1 \times V_1 = 2.0 \times 3.0 = 6.0 \text{ L atm} \]The final state has pressure \( P_2 = 4.0 \) atm and volume \( V_2 = 5.0 \) L, so the final \( PV \) value is:\[ (PV)_2 = P_2 \times V_2 = 4.0 \times 5.0 = 20.0 \text{ L atm} \]
3Step 3: Calculating Change in PV
The change in the \( PV \) term is the difference between the final and initial \( PV \) values:\[ \Delta (PV) = (PV)_2 - (PV)_1 = 20.0 - 6.0 = 14.0 \text{ L atm} \]
4Step 4: Calculating Change in Enthalpy \( \Delta H \)
Substitute the known values for \( \Delta U \) and \( \Delta (PV) \) into the enthalpy equation from Step 1:\[ \Delta H = \Delta U + \Delta (PV) = 30.0 + 14.0 = 44.0 \text{ L atm} \]
5Step 5: Choosing the Correct Answer from Given Options
The calculated change in enthalpy \( \Delta H = 44.0 \text{ L atm} \). According to the provided options, the correct answer is: (c) 44.0
Key Concepts
Internal EnergyPV WorkNon-Ideal Gas Behavior
Internal Energy
Internal energy is a fundamental concept in thermodynamics, representing the total energy contained within a system. It encompasses both the kinetic and potential energy of the particles that make up the system. Changes in internal energy can occur due to heat exchange or work done by or on the system.
In our exercise, we are given the change in internal energy, \( \Delta U = 30.0 \, \text{L atm} \). This indicates how much energy the system’s particles have gained or lost during the process. Understanding how internal energy changes are crucial for analyzing thermodynamic processes like those of gases.
When calculating changes in enthalpy (\( \Delta H \)), the change in internal energy serves as the basis for further computations, stressing its pivotal role in energy considerations.
In our exercise, we are given the change in internal energy, \( \Delta U = 30.0 \, \text{L atm} \). This indicates how much energy the system’s particles have gained or lost during the process. Understanding how internal energy changes are crucial for analyzing thermodynamic processes like those of gases.
When calculating changes in enthalpy (\( \Delta H \)), the change in internal energy serves as the basis for further computations, stressing its pivotal role in energy considerations.
PV Work
PV work, or pressure-volume work, is an essential component in the study of thermodynamics. It describes the work done by or on a system as it expands or compresses against external pressure. In an ideal gas, the work done is directly proportional to changes in its volume while the pressure is held constant. However, in non-ideal gases like those in this exercise, the relationship may vary.
To calculate PV work, we look at the product of the initial and final pressures and volumes. For example, the change in the PV product, \( \Delta (PV) \), is calculated as the difference between the final and initial states. Here, the initial product was \( 6.0 \, \text{L atm} \) and final product \( 20.0 \, \text{L atm} \), giving a change of \( \Delta (PV) = 14.0 \, \text{L atm} \).
Understanding PV work is vital as it forms part of the "flow work" in open systems, affecting how energy is transferred within and out of a system. It’s an integral part of determining enthalpy changes in processes.
To calculate PV work, we look at the product of the initial and final pressures and volumes. For example, the change in the PV product, \( \Delta (PV) \), is calculated as the difference between the final and initial states. Here, the initial product was \( 6.0 \, \text{L atm} \) and final product \( 20.0 \, \text{L atm} \), giving a change of \( \Delta (PV) = 14.0 \, \text{L atm} \).
Understanding PV work is vital as it forms part of the "flow work" in open systems, affecting how energy is transferred within and out of a system. It’s an integral part of determining enthalpy changes in processes.
Non-Ideal Gas Behavior
Non-ideal gas behavior refers to the deviation of real gases from the ideal gas laws due to factors like intermolecular forces and the volume occupied by gas molecules. While ideal gas laws provide a good approximation of behavior under many conditions, real gases often exhibit complex patterns, especially under high pressure or low temperature.
In our exercise, the behavior of a non-ideal gas is examined, which requires more sophisticated understanding compared to ideal gases. Such deviations are often corrected using factors or equations like the Van der Waals equation or compressibility factors, which adjust the ideal gas law to more accurately reflect observed behaviors.
Recognizing non-ideal gas behavior in calculations can give students and practitioners a better ground for making accurate predictions about gas interactions, ensuring that the calculations of things like changes in enthalpy remain valid under realistic conditions. Understanding these deviations is critical for chemical engineering, environmental science, and other fields dealing with gas processing and applications.
In our exercise, the behavior of a non-ideal gas is examined, which requires more sophisticated understanding compared to ideal gases. Such deviations are often corrected using factors or equations like the Van der Waals equation or compressibility factors, which adjust the ideal gas law to more accurately reflect observed behaviors.
Recognizing non-ideal gas behavior in calculations can give students and practitioners a better ground for making accurate predictions about gas interactions, ensuring that the calculations of things like changes in enthalpy remain valid under realistic conditions. Understanding these deviations is critical for chemical engineering, environmental science, and other fields dealing with gas processing and applications.
Other exercises in this chapter
Problem 19
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