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The internal energy change when a system goes from state \(\mathrm{A}\) to \(\mathrm{B}\) is \(40 \mathrm{~kJ} / \mathrm{mol}\). If the system goes from \(\mathrm{A}\) to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy? (a) \(40 \mathrm{~kJ}\) (b) \(>40 \mathrm{~kJ}\) (c) \(<40 \mathrm{~kJ}\) (d) zero

Short Answer

Expert verified
(d) zero

Step by step solution

01

Understanding Internal Energy Change

Internal energy is a state function, meaning it depends only on the initial and final states of the system, not on the path taken to get there. Therefore, if a system starts at state A, moves to state B, and then returns to state A, the net change in internal energy is determined by the state function property.
02

Calculating Net Internal Energy Change

When the system goes from state A to state B, the change in internal energy is given as 40 kJ/mol. Since internal energy is a state function, returning from state B to state A, regardless of the reversibility of the path, will result in the energy change between state A and state B being zero over the entire cycle. Therefore, the net internal energy change is zero.

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Key Concepts

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

Internal Energy
Internal energy is a key concept in thermodynamics. It represents the total energy contained within a system. This energy consists of both kinetic energy, due to the movement of molecules, and potential energy, arising from the interactions between molecules.
There are a few important characteristics of internal energy that are crucial for understanding problems like the one presented:
  • Internal energy is an extensive property, meaning it depends on the amount of substance present in the system.
  • It is a state function, which implies that the change in internal energy depends only on the initial and final states, not the path taken to move between these states.
  • In a closed system, changes in internal energy can occur through heat transfer or performance of work.
Understanding these facets helps us make sense of thermodynamic processes and calculate changes in internal energy.
State Functions
State functions are properties of a system that depend only on the current state of the system, not how it reached there. This is an essential concept in thermodynamics because it allows simplification of complex problems.
Some critical aspects include:
  • Examples of state functions include internal energy, enthalpy, entropy, and Gibbs free energy.
  • State functions enable us to evaluate the change in properties between two states simply by subtracting the initial state value from the final state value.
  • Since they do not depend on the path, knowing initial and final states is sufficient to understand the change in a state function.
In the problem where a system returns to its original state, knowing that internal energy is a state function allows us to conclude that the net change is zero.
Reversible and Irreversible Processes
Understanding reversible and irreversible processes is vital in the study of thermodynamics. Reversible processes are idealized processes that occur infinitely slowly, allowing the system to adjust and remain in equilibrium with its surroundings at each step.
Irreversible processes are more real-world processes that occur quickly, creating some level of irreversibility due to factors like friction or rapid expansion.
  • In reversible processes, you can restore both the system and the surroundings to their original states without any extra energy.
  • In irreversible processes, achieving the original states requires external intervention and often expends additional energy.
  • Irreversible processes are more prevalent in natural phenomena because they happen faster and are influenced by uncontrolled external conditions.
In the given exercise, despite taking different paths (one reversible, one irreversible), the net change in internal energy remains zero due to its nature as a state function.

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Most popular questions from this chapter

\(15 \mathrm{~mL}\) of gaseous hydrocarbon requires \(45 \mathrm{~mL}\) of oxygen for complete combustion which produces \(30 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) gas, measured under identical conditions. The formula of the hydrocarbon is \(\mathrm{C}_{x} \mathrm{H}_{y}\). The ratio \(\underline{\mathrm{y}}\) is X

Which of the following equations represent standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{4} ?\) (a) \(2 \mathrm{C}\) (diamond) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (b) \(2 \mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (c) \(2 \mathrm{C}\) (diamond) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (d) \(2 \mathrm{C}\) (graphite) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

If a gas at constant temperature and pressure expands, then its (a) internal energy decreases (b) entropy increases and then decreases (c) internal energy increases (d) internal energy remains constant

The standard entropies of \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{C}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) are \(213.5,5.74\) and \(205 \mathrm{JK}^{-1}\) respectively. The standard entropy of the formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(1.16 \mathrm{~J} \mathrm{~K}^{-1}\) (b) \(2.76 \mathrm{~J} \mathrm{~K}^{-1}\) (c) \(1.86 \mathrm{~J} \mathrm{~K}^{-1}\) (d) \(2.12 \mathrm{~J} \mathrm{~K}^{-1}\)

For complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \ell+\) \(3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O} \ell\) the amount of heat produced as measured in bomb calorimeter, is \(1364.47 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming ideality the Enthalpy of combustion, \(\Delta \mathrm{H}\) for the reaction will be: \(\left(\mathrm{R}=8.314 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) \(-1460.50 \mathrm{kj} \mathrm{mol}^{-1}\) (b) \(-1350.50 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-1366.95 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1361.95 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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