JEE Mains · Physics · STD 11 - 11. thermodynamics
Three different processes that can occur in an ideal monoatomic gas are shown in the \(P\) vs \(V\) diagram. The paths are labelled as \(A \rightarrow B, A \rightarrow C\) and \(A \rightarrow D .\) The change in internal energies during these process are taken as \(E _{ AB }, E _{ AC }\) and \(E _{ AD }\) and the work done as \(W _{ AB }\) \(W _{ AC }\) and \(W _{ AD }\) The correct relation between these parameters are

- A \(E _{ AB }= E _{ AC }= E _{ AD }, W _{ AB }>0, W _{ AC }=0,\) \(W _{ AD }>0\)
- B \(E _{ AB }< E _{ AC }< E _{ AD }, W _{ AB }>0, W _{ AC }> W _{ AD }\)
- C \(E _{ AB }= E _{ AC }< E _{ AD }, W _{ AB }>0, W _{ AC }=0\) \(W _{ AD }<0\)
- D \(E _{ AB }> E _{ AC }> E _{ AD }, W _{ AB }< W _{ AC }< W _{ AD }\)
Answer & Solution
Correct Answer
(A) \(E _{ AB }= E _{ AC }= E _{ AD }, W _{ AB }>0, W _{ AC }=0,\) \(W _{ AD }>0\)
Step-by-step Solution
Detailed explanation
\(\Delta U = nC _{ v } \Delta T =\) same \(AB \rightarrow\) volume is increasing \(\Rightarrow W >0\) \(AD \rightarrow\) volume is decreasing \(\Rightarrow W <0\) \(AC \rightarrow\) volume is constant \(\Rightarrow W =0\)
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