Question

a system absorbs 300 calorie of heat with the result of that , the volume of system becomes double of its initial volume and temp changes from 273k to 546k. the work done by the system on he sorroundings is 200 calorie. calculate change in internal energy:

Answer 1

Dear Student:

Relation between ΔQ,ΔU and  ΔW is given as

ΔQ=ΔU + ΔW

Where,ΔQ = change in heat=300 calorie

           ΔU = Change in internal energy=?

           ΔW= work done=200 calorie.

         ΔQ-ΔW=ΔU

         300-200=ΔU

                 ΔU = 100 calorie.

Hence,internal energy change is 100 calorie.

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