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One mole of an non linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to 1 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 `W=DeltaU=C_(v)(T_(2)-T_(1))` `C_(v)` for triatomic NON linear GAS = 3R (i) For rev. process. `P_(1)^(1-GAMMA)T_(1)^(gamma)=P_(2)^(1-gamma)T_(2)^(gamma)""r=4/3` `T_(2)=T_(1)(P_(1)/P_(2))^((1-gamma)/v)=300(16)(1-4/3)/(4//3)=300(2^(4))^(-1//4)=150K` `DeltaU=w-3R(150-300)=-450R` (II) n = 1 `-P_("ext")(V_(2)-V_(1))=C_(V)(T_(2)-T_(1))` `-P_("ext")((RT_(2))/P_(2)-(RT_(1))/P_(1))=3R(T_(2)-T_(1))` `-1((RT_(2))/1-(RT_(1))/16)=3R(T_(2)-T_(1))` `-R/16(16T_(2)-T_(1))=3R(T_(2)-T_(1))` `-16T_(2)+T_(1)=48T_(2)-48T_(1)` `49T_(1)=64T_(2)` `T_(2)=49/64xx300` `T_(2)=229.69` `W_("irr")+C_(v)(T_(2)-T_(1))+3R(229.69-300)=-210.93R` |
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