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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. Find the values of DeltaS_("sys"),DeltaS_("surr")&DeltaS_("tota") under the following conditions. (i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly (iii) Expansion is free. |
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Answer» Solution :For non-linear tri-atomic IDEAL gas `C_(v)=3R,C_(p)=4R` (i) `DeltaS_("sys")=nC_(v)"ln"T_(2)/T_(1)+NR"ln"v_(2)/v_(1)=0` q = 0 `DeltaS_("surr")=-DeltaS_("sys")=0` `DeltaS_("total")=0` (ii) First of all we will have to calculate the temperature of the gas after it has undergoes the said adiabatic reversible expansion we have q = 0 `DeltaU=q+w` `nC_(v)(T_(2)-T_(1))=-P_("ext")(v_(2)-v_(1))` `3R(T_(2)-300)=-1[(RT_(2))/p_(2)-(RT_(1))/2]=-R[T_(2)/1-300/16]` `T_(2)=229.68K` `DeltaS_("sys")=nC_(p)"ln"T_(2)/T_(1)+nR"ln"p_(1)/p_(2)` `=4R"ln"229.68/300+R"ln"16/1=-1.068R+2.77` `R=1.702R` `DeltaS_("surr")=(-q_("irr"))/T=0` `DeltaS_("total")=DeltaS_("sys")=1.702R` (iii) In free adiabatic expansion we have w = 0 `q=0" "DeltaT=0` `DeltaS_("sys")=nR"ln"p_(1)/p_(2)=Rln16=2.77R` `DeltaS_("surr")=(-q_("irr"))/T=0` `DeltaS_("total")=DeltaS_("sys")=2.77R` |
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