If one mole of an ideal gas at (P1 V1) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one- half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B ® C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to:

Option 1 - <p>0</p>
Option 2 - <p>RT In2</p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>R</mi> <mi>T</mi> <mrow> <mo>(</mo> <mrow> <mi>I</mi> <mi>n</mi> <mn>2</mn> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mrow> <mi>γ</mi> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mo>−</mo> <mfrac> <mrow> <mi>R</mi> <mi>T</mi> </mrow> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mrow> <mi>γ</mi> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </math> </span></p>
5 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 3
Detailed Solution:

n = 1 mole

W A B = n R T l n v 2 v 1 = 1 * R T l n 2 v 1 v 1 = R T l n 2

W B C = 0

W C A = P 1 V 1 P 2 V 2 γ 1 = P 1 4 * 2 V 1 P 1 * V 1 γ

W C A = R T 2 ( γ 1 )

W t o t a l = R T [ l n 2 1 2 ( γ 1 ) ]

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Physics Thermodynamics 2025

Physics Thermodynamics 2025

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