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When heat energy of 1500J is supplied to a gas the external workdone by the gas is 525J what is the increase in its internal energy |
Answer» Heat energy supplied `DeltaQ=1500J` External workdone `DeltaW=525J` By `1^(st)` law of thermodynamics `DeltaQ=DeltaU+DeltaW` `therefore DeltaU=DeltaQ-DeltaW=1500-525=975J`. |
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