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In an isobaric process (`lambda` is adiabatic exponent of the gas)A. The heat given to gas is `(lambda)/(lambda-1)` times the work done by gas.B. The work done by gas is `(lambda-1)` time the change in internal energy.C. The temperature of gas in increased.D. The temperature of gas is decreased. |
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Answer» At constant pressure. If volume increases. Temperature also increases volume decreases, In isobaric process, `DeltaQ=lambda.DeltaU` `therefore DeltaQ=DeltaU+DeltaW` `therefore DeltaW=lambda.DeltaU-DeltaU=(y-1)DeltaU` `DeltaW=(y-1)DeltaU` `thereforeDeltaQ=(lambda)/((y-1))DeltaW` |
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