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When a system is taken from i to state f along the path iaf, it is found that Q = 50 cal and W =20 cal. Along the path ibf Q= 36 cal . W along the path ibf is

  • Option 1)

    14 cal

  • Option 2)

    6 cal

  • Option 3)

    16 cal

  • Option 4)

    66 cal

 

Answers (1)

best_answer

As we learnt in 

Change in internal energy for cyclic process -

\Delta U= 0
 

- wherein

Since in a cyclic process initial and final state is same.

U_{f}= U_{i}

 

 Change in internal energy is same for both path.

\Delta U_{iaf}\:=\:\Delta U_{ibf}

=> \:(50-20) \:Cal\:=(36\:Cal-W_{ibf})

=> W_{ibf}\:=6\:Cal


Option 1)

14 cal

This option is incorrect.

Option 2)

6 cal

This option is correct.

Option 3)

16 cal

This option is incorrect.

Option 4)

66 cal

This option is incorrect.

Posted by

Aadil

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