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Which one of the following statements applicable to a perfect gas will also be true for an irreversible process
- a)δQ = dU + PdV
- b)δQ = TdS
- c)TdS = dU + PdV
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Which one of the following statements applicable to a perfect gas will...
δQ = dU+ PdV... Applicable for a closed system when only PdV work is present. This is true only for a reversible process.
δQ = TdS ...Applicable for a reversible process.
TdS = dU + PdV... Applicable for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path.
δQ = TdS ...Applicable for a reversible process.
TdS = dU + PdV... Applicable for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path.
Most Upvoted Answer
Which one of the following statements applicable to a perfect gas will...
Understanding the Statement
In thermodynamics, the behavior of a perfect gas during processes is well defined. The statement in question pertains to the first law of thermodynamics and its application during irreversible processes.
Irreversible Process Characteristics
- An irreversible process is one that cannot be reversed without leaving a net change in the system and surroundings.
- These processes often involve friction, turbulence, or other dissipative effects.
Key Thermodynamic Equations
1. First Law of Thermodynamics
- The first law states that the change in internal energy (dU) is equal to the heat added to the system (delta Q) minus the work done by the system (PdV).
- For a perfect gas, this can be expressed as:
delta Q = dU + PdV
2. Entropy Changes
- For any process, the change in entropy (dS) relates to heat transfer (delta Q) and temperature (T) as:
delta Q = TdS
- This relationship holds true for both reversible and irreversible processes.
Correct Statement Analysis
- The statement TdS = dU + PdV is derived from combining the first law of thermodynamics with the definition of entropy.
- This equation is valid regardless of whether the process is reversible or irreversible, as it reflects the fundamental principles of thermodynamics.
Conclusion
Thus, in the context of the question, option 'C' (TdS = dU + PdV) is indeed correct. This relationship is universally applicable to any process involving a perfect gas, highlighting the interplay between energy, heat, and entropy, even in irreversible processes.
In thermodynamics, the behavior of a perfect gas during processes is well defined. The statement in question pertains to the first law of thermodynamics and its application during irreversible processes.
Irreversible Process Characteristics
- An irreversible process is one that cannot be reversed without leaving a net change in the system and surroundings.
- These processes often involve friction, turbulence, or other dissipative effects.
Key Thermodynamic Equations
1. First Law of Thermodynamics
- The first law states that the change in internal energy (dU) is equal to the heat added to the system (delta Q) minus the work done by the system (PdV).
- For a perfect gas, this can be expressed as:
delta Q = dU + PdV
2. Entropy Changes
- For any process, the change in entropy (dS) relates to heat transfer (delta Q) and temperature (T) as:
delta Q = TdS
- This relationship holds true for both reversible and irreversible processes.
Correct Statement Analysis
- The statement TdS = dU + PdV is derived from combining the first law of thermodynamics with the definition of entropy.
- This equation is valid regardless of whether the process is reversible or irreversible, as it reflects the fundamental principles of thermodynamics.
Conclusion
Thus, in the context of the question, option 'C' (TdS = dU + PdV) is indeed correct. This relationship is universally applicable to any process involving a perfect gas, highlighting the interplay between energy, heat, and entropy, even in irreversible processes.
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Which one of the following statements applicable to a perfect gas will also be true for an irreversible processa)δQ = dU + PdVb)δQ = TdSc)TdS =dU + PdVd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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