InterviewSolution
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InterviewSolution
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One mole of an ideal mono-atomic gas is taken round cyclic process `ABCD` as shown in figure below. Calculate |
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Answer» Path `AB` is isochoric `(w_(1)=0)`, path `BC` is isothermal `(w_(2)=-ve)`, path `CA` is isobaric `(w_(3)=+ve)` Total work done by gas `(w)` `=w_(1)+w_(2)+w_(3)` `=0+2.303nRT"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))` `=0+2.303P_(B)V_(B)"log"(V_(B))/(V_(C))+P(V_(C)-V_(A))` `=2.303xx3P_(0)V_(0)"log"(V_(0))/(2V_(0))+P_(0)(2V_(0)-V_(0))` `=-2P_(0)V_(0)+P_(0)V_(0)=-P_(0)V_(0)` Also, `w_(2)=-2P_(0)V_(0)` and `w_(3)=P_(0)V_(0)` also, For the path `AB`, ie. isochoric `q_(1)=nxxC_(P)xx(T_(B)-T_(A))=1xx3/2R[(P_(B)V_(B)-P_(A)V_(A))/R]` `=3/2[3P_(0)V_(0)-P_(0)V_(0)]=+3P_(0)V_(0)` For the path `CA`, i.e., isochoric: `q_(3)=nxxC_(P)xx(T_(A)-T_(B))` `=1xx5/2R[(P_(A)V_(A)-P_(B)V_(B))/R]` `=5/2[P_(0)V_(0)-2P_(0)V_(0)]` `q_(3)=-5/2 P_(0)V_(0)` Also net heat absorbed `=3P_(0)V_(0)-5/2P_(0)V_(0)-=(P_(0)V_(0))/2` `:. q_("net")=(P_(0)V_(0))/2` Also, `(P_(0)V_(0))/(T_(1))=(3P_(0)V_(0))/(T_(2)):. T_(2)=3T_(1)=(3P_(0)V_(0))/R` |
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