In the reported figure, there is a cyclic process ABCDA on a sample of 1mol of a diatomic gas. The temperature of the gas during the process A→B and C→D are T? and T? (T? > T?) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

Option 1 - <p>W_AB = W_DC<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>W_AD = W_BC<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>W_BC &lt; W_AD<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>W_BC + W_DA &gt; 0</p>
9 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 3
Detailed Solution:

A→B & D→C (isothermal process)
So, TA = TB & TD = TC. Now B→C & D→A (adiabatic process)
|WBC| = nR/ (γ-1) (TB - TC)
|WAD| = nR/ (γ-1) (TA - TD) = nR/ (γ-1) (TB - TC)
∴ |WBC| = |WAD|

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Physics Thermodynamics 2025

Physics Thermodynamics 2025

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