1.

When heat energy of 1500 J is suppliedto a gas the external work done on the gas is 525 J what is the increase in its internal energy

Answer»

Solution :Heat energy SUPPLIED `DELTAQ= +1500J`
EXTERNAL work done `Deltaw=-525J`
By `1^(st)` law of THERMODYNAMICS `DeltaQ = DeltaU +Deltaw`
`:. DeltaU =DeltaQ-Deltaw=1500+525=1025J`


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