Discussion Forum

1)

 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 path af, ib and bf are  W_af=200J, W_ib=50J and  W_bf= 100 J respectively. The heat supplied to the system along the path iaf,ib and bf are Q_iaf, Q_ib and 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 

Answer:

Option C

Explanation:

  $W_{ibf}=W_{ib}+W_{bf}$

   =50J +100J=150J

$W_{iaf}=W_{ia}+W_{af}$

  =0+200 J= 200J

  Q_iaf= 500J

  So,   $\triangle U_{iaf}=Q_{iaf}-W_{iaf}$

                    =500 J-200J=300J

           =U_f-U_i

 So,     $U_{f}=U_{iaf}+U_{i}$

            =300 J+100J

              =400J

 $\triangle U_{ib}=U_{b}-U_{i}$

            =200J-100J

              =100 J

 $Q_{ib}=\triangle U_{ib}+W_{ib}$

             =100J +50J=150J

$Q_{ibf}=\triangle U_{ibf}+W_{ibf}$

   $=\triangle U_{iaf}+W_{ibf}$

        =300J+150J

      =450 J

So, the required ratio,

 $\frac{Q_{bf}}{Q_{ib}}=\frac{Q_{ibf}-Q_{ib}}{Q_{ib}}$

    =  $\frac{450-150}{150}=2$