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A thermodynamic system is taken from an initial state I with internal energy `U_i=-100J` to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the pat af, ib and bf are `W_(af)=200J, W_(ib)=50J and W_(bf)=100J` respectively. The heat supplied to the system along the path iaf, ib and bf are `Q_(iaf), Q_(ib),Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b=200J and Q_(iaf)=500J`, The ratio `(Q_(bf))/(Q_(ib))` is |
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Answer» Correct Answer - B Here, `Q_(iaf)= 500 Jand W_(af)= W_(iaf)= 200J` `:. U_(iaf)= U_(f)-U_(i)= Q_(iaf)-W_(iaf)` `= 500-200= 300J` `:. U_(f)= 300+U_(i)= 300+100= 400 J` Now, `Q_(ib)= U_(ib)+W_(ib)= (U_(b)-U_(i))+W_(ib)` `=(200-100)+50= 150J` and `Q_(bf)=U_(bf)+W_(bf)=(U_(f)-U_(b))+W_(bf)` `=(400-200)+100= 300 J` `:. (Q_(bf))/(Q_(ib))=(300)/(150)=2` |
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