question_answer

Figure below shows two paths that may be taken by a gas to go from a state A to a state C.         In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be                                            

A)  380 J                    

B)  500 J    

C)         460 J              

D)  300 J

Correct Answer: C

Solution :

                Since, initial and final points are same So,  \[\Delta {{U}_{A\to B\to C}}=\Delta {{U}_{A\to C}}\]                 ?(i) Also \[A\to B\] is isochoric process So     \[d{{W}_{A\to B}}=0\] and      \[dQ=dU+dW\] So,     \[d{{Q}_{A\to B}}=d{{U}_{A\to B}}=400J\] Next \[B\to C\] is isobaric process So,      \[d{{O}_{B\to C}}=d{{U}_{B\to C}}+d{{W}_{B\to C}}\]                 \[=d{{U}_{B\to C}}+p\Delta {{V}_{B\to C}}\] \[\Rightarrow \,\,100=d{{U}_{B\to C}}+6\times {{10}^{4}}(2\times {{10}^{-3}})\] \[\Rightarrow \,\,D{{U}_{B\to C}}=100-120=-20J\] From Eq. (i), \[\because \,\,\,\,\,\Delta {{U}_{A\to B\to C}}=\Delta {{U}_{A\to C}}\] \[\Rightarrow \,\,\,\Delta {{U}_{A\to B}}+\Delta {{U}_{B\to C}}=d{{Q}_{A\to C}}-d{{W}_{A\to C}}\] \[\Rightarrow \,\,\,400+\left( -20 \right)=d{{Q}_{A\to C}}\]                                 \[-(p\Delta {{V}_{A}}+Area\,of\,\Delta ABC)\] \[\Rightarrow \,\,d{{Q}_{A\to C}}=380+\left( \begin{align}   & 2\times {{10}^{4}}\times 2\times {{10}^{-3}} \\  & +\frac{1}{2}\times 2\times {{10}^{-3}}\times 4\times {{10}^{4}} \\ \end{align} \right)\]                                 \[=380+(40+40)\]                 \[d{{Q}_{A\to C}}=460J\]