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One mole of an non linear triatomic ideal gas is compressed adiabatically at 300 K from 1 atom to 16 atm. Calculate Work done under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly. |
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Answer» SOLUTION :q = 0 Adiabatic PROCESS (i) `W_("rev")=DeltaU=C_(V)(T_(2)-T_(1))` `P^(1-gamma)T^(gamma)=K` `T_(2)=300(1/16)^((1-4//3)/(4//3))=600K` w = 3R (600 - 300) = 900 R (ii) `W_("irr")=-P_("EXT")(V_(2)-V_(1))=-P_("ext")((nRT_(2))/P_(2)-(nRT_(1))/P_(1))` `-P_("ext")((nRT_(2))/P_(2)-(nRT_(1))/P_(1))=C_(v)(T_(2)-T_(1))` `-16((RT_(2))/16-(RT_(1))/1)=3R(T_(2)-T_(1))` `-(T_(2)-16T_(1))=3R(T_(2)-T_(1))` `-T_(2)+16T_(1)=3T_(2)-3T_(1)` `4T_(2)=19T_(1)` `T_(2)=19/4xx300=1425K` `w=DeltaU=3R(1425-300)=3375R` |
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