Question

A thermodynamic system is taken from an initial state i with internal energy $U_i=100$J 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 paths af, ib and bf are $W_{af}=200$J, $W_{ib}=50$J 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 $U_b=200$J and $Q_{iaf}=500$J, the ratio $Q_{bf}/Q_{ib}$ is?

Answer 1

solution:

$Q_{iab}$= 500 j and $W_{af}$ = $W_{iaf}$= 200J

$\therefore$ $U_{iaf}$ = $U_f-U_i=Q_{iaf}-W_{iaf}$

$\to$ 500 - 200 =300 J

$U_f=300+U_i$ =400 J

$Q_{ib}=U_{ib}+W_{ib}=(U_b-U_i)+W_{ib}$

$\to$ 200-100 + 50 = 150J

$Q_{bf}=U_{bf}+W_{bf}=(U_f-U_b)+W_{bf}$

$\to$(400−200)+100=300J

$\frac{Q_{bf}}{Q_{ib}}=\frac{300}{150}=2$

hence the correct answer is 2