A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

 

Option 1 -

20 J

Option 2 -

30 J

Option 3 -

-30 J

Option 4 -

 -60J

0 4 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    Given                                                                           

    QAB = +40J

    QBC = 0J

    QCA = -60J

    WCA =?

    WBC = -50J (on the gas)

    UA = 1560J

    1st Law on path A to B

    WAB +   Δ U = Q

    Þ 0 +    [ U B U A ] = 4 0

    Þ UB = UA + 40

    = 1560 + 40

    UB = 1600 J

     1st Low

    ...more

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