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  2. Please help! - Chemical Thermodynamics - JEE Main
# Engineering

During compression of a spring the work done is 10\:kJ\:\:and\:\:2\:kJ  escaped to the surrounding as heat . The change in internal energy , \Delta U(in\:\:kJ)   is :

  • Option 1)

    -12

  • Option 2)

    -8

  • Option 3)

    8

  • Option 4)

    12

Answers (1)
S solutionqc

 

Internal Energy -

Fo an ideal gas E is a function of temperature only

\Delta E= K.E_{Translational}+K.E_{Rotational }+K.E_{vibrational}

- wherein

With change in temperature only kinetic energy is changed but not any potential energy.

 

 

First law of Thermodynamics -

Energy of universe is always conserved or total energy of an isolated system is always conserved

\Delta E= q + W


 

- wherein

\Delta E= Internal Energy

q= Heat

W= work

 

 

 

 

we know ,

\\ \Delta U=q+w\\\\\:w=10\:\:kJ(positive\:\:compression)

\\ q=-2\:KJ(negative,\:heat\:\:released/escaped)\\\\\:\Delta V=-2+10=8\:\:KJ


Option 1)

-12

Option 2)

-8

Option 3)

8

Option 4)

12

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