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At 27^(@)C , two moles of an ideal mono-atomic gas occupy a volumeV. The gas expands adiaba-tically to a volume 2V. Calculate (a) final temperature of the gas (b) Change in its internal energy and (c ) the work done by the gas during the process. |
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Answer» Solution :(a) In CASE of ADIABATIC change, `PV^(Gamma)` = const. with PV=nRT So that `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) ["with " gamma = (5//3)]` i.e., `300xxV^(2//3) = T(2V)^(2//3) or T=300//(2)^(2//3)=189K` (b) As `Delta U=(2xx8.31xx(189-300))/([5/3-1]) = -2767.23 J` NEGATIVE sign means internal energy decreases. (c) According to first law of thermodynamics, `DeltaQ=DeltaU+DeltaW` and as for adiabatic change `DeltaQ=0 rArr DeltaW = -DeltaU=2767.23J` . |
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