### Neet Solved Paper 2015 Question 19

##### Question: 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

#### Options:

A) 380 J

B) 500 J

C) 460 J

D) 300 J

## Show Answer

#### Answer:

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+( -20 )=d{Q _{A\to C}} $

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

$ \Rightarrow d{Q _{A\to C}}=380+( \begin{aligned} & 2\times 10^{4}\times 2\times {10^{-3}} \\ & +\frac{1}{2}\times 2\times {10^{-3}}\times 4\times 10^{4} \\ \end{aligned} ) $

$ =380+(40+40) $ $ d{Q _{A\to C}}=460J $