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), \[\because \,\,\,\,\,\Delta {{U}_{A\to B\to C}}=\Delta {{U}_{A\to C}}\] \[\Rightarrow \,\,\,\Delta {{U}_{A\to B}}+\Delta {{U}_{B\to C}}=d{{Q}_{A\to C}}-d{{W}_{A\to C}}\] \[\Rightarrow \,\,\,400+\left( -20 \right)=d{{Q}_{A\to C}}\] \[-(p\Delta {{V}_{A}}+Area\,of\,\Delta ABC)\] \[\Rightarrow \,\,d{{Q}_{A\to C}}=380+\left( \begin{align} & 2\times {{10}^{4}}\times 2\times {{10}^{-3}} \\ & +\frac{1}{2}\times 2\times {{10}^{-3}}\times 4\times {{10}^{4}} \\ \end{align} \right)\] \[=380+(40+40)\] \[d{{Q}_{A\to C}}=460J\]You need to login to perform this action.
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