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A system goes from P to Q by two different paths in the P-V diagram as shown in figure. Heat given to the system in path 1 is 1000 J. - Physics

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प्रश्न

A system goes from P to Q by two different paths in the P-V diagram as shown in figure. Heat given to the system in path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

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उत्तर

According to the first law of thermodynamics,

∆Q = AU + ∆W.

Let us apply this for each path.

For path 1: Heat given Q1 = + 1000 J

Let work done for path 1 = W1

For path 2: Work done (W2) = (W1 – 100) J

Heat has given Q2 = ?

A change in internal energy between two states for the different paths is the same.

∴ ∆U = Q1 – W1 = Q2 – W2

1000 – W1 = Q2 – (W1 – 100)

⇒ Q2 = 1000 – 100 = 900 J

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First Law of Thermodynamics
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पाठ 12: Thermodynamics - Exercises [पृष्ठ ८७]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
पाठ 12 Thermodynamics
Exercises | Q 12.13 | पृष्ठ ८७

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