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                                                                           [NEET  2015]

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),

\[\because \]       \[\Delta {{U}_{A\to B\to C}}=\Delta {{U}_{A\to C}}\]

\[\Rightarrow \]   \[\Delta {{U}_{A\to B}}+\Delta {{U}_{B\to C}}=d{{Q}_{A\to C}}-d{{W}_{A\to C}}\]

\[\Rightarrow \]   \[400+\left( -20 \right)=d{{Q}_{A\to C}}\]

\[-(p\Delta {{V}_{A}}+Area\,of\,\Delta ABC)\]

\[\Rightarrow \]       \[d{{Q}_{A\to C}}=380+\left( \begin{align}   & 2\times {{10}^{4}}\times 2\times {{10}^{-3}} \\  & +\frac{1}{2}\times 2\times {{10}^{-3}}\times 4\times {{10}^{4}} \\ \end{align} \right)\]

\[=380+(40+40)\]

\[d{{Q}_{A\to C}}=460J\]