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Which of the following relationships is valid only
for reversible processes undergone by a closedsystem of simple compressible substance (neglect changes in kinetic and potential energy) ?
a) δQ = dU + δW
b) TdS = dU + pdV
c) TDS = dU + δW
d) δQ = dU + pdV
Correct answer is option 'D'. Can you explain this answer?
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Which of the following relationships is valid only... more for reversi...
This question is part of UPSC exam. View all Mechanical Engineering courses
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Which of the following relationships is valid only... more for reversi...
Explanation:
Reversible Processes in a Closed System:
In reversible processes undergone by a closed system of a simple compressible substance, the changes in kinetic and potential energy are neglected. This means that the system is assumed to be in a state of thermodynamic equilibrium throughout the process.
Validity of Relationship:
The relationship δQ = dU + pdV is valid only for reversible processes in a closed system of a simple compressible substance. Here's why:
- δQ represents the heat transfer to the system.
- dU is the change in internal energy of the system.
- pdV represents the boundary work done on the system.
Explanation of Options:
a) δQ = dU + δW: This relationship is not valid for reversible processes as it includes δW, which may not be equal to pdV in reversible processes.
b) TdS = dU + pdV: This relationship involves entropy change and temperature, which may not hold true for all reversible processes.
c) TDS = dU + δW: This relationship includes δW, which is not specific to reversible processes.
Correct Answer:
d) δQ = dU + pdV: This relationship is valid for reversible processes undergone by a closed system of a simple compressible substance, making it the correct choice for this scenario.
Reversible Processes in a Closed System:
In reversible processes undergone by a closed system of a simple compressible substance, the changes in kinetic and potential energy are neglected. This means that the system is assumed to be in a state of thermodynamic equilibrium throughout the process.
Validity of Relationship:
The relationship δQ = dU + pdV is valid only for reversible processes in a closed system of a simple compressible substance. Here's why:
- δQ represents the heat transfer to the system.
- dU is the change in internal energy of the system.
- pdV represents the boundary work done on the system.
Explanation of Options:
a) δQ = dU + δW: This relationship is not valid for reversible processes as it includes δW, which may not be equal to pdV in reversible processes.
b) TdS = dU + pdV: This relationship involves entropy change and temperature, which may not hold true for all reversible processes.
c) TDS = dU + δW: This relationship includes δW, which is not specific to reversible processes.
Correct Answer:
d) δQ = dU + pdV: This relationship is valid for reversible processes undergone by a closed system of a simple compressible substance, making it the correct choice for this scenario.
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Which of the following relationships is valid only... more for reversible processes undergone by a closedsystem of simple compressible substance (neglect changes in kinetic and potential energy) ?a) δQ = dU + δWb) TdS = dU + pdVc) TDS = dU + δWd) δQ = dU + pdVCorrect answer is option 'D'. Can you explain this answer?
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