JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
An ideal gas undergoes a cyclic transformation starting from the point \(A\) and coming back to the same point by tracing the path \(\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow \mathrm{A}\) as shown in the three cases above.
Choose the correct option regarding \(\Delta \mathrm{U}\) :
- A \(\Delta \mathrm{U}(\) Case-I \()=\Delta \mathrm{U}(\) Case-II \()=\Delta \mathrm{U}(\) Case-III \()\)
- B \(\Delta \mathrm{U}(\) Case-I \()\gt\Delta \mathrm{U}(\) Case-III \()\gt\Delta \mathrm{U}(\) Case-II \()\)
- C \(\Delta \mathrm{U}(\) Case-III \()\gt\Delta \mathrm{U}(\) Case-II \()\gt\Delta \mathrm{U}(\) Case-I \()\)
- D \(\Delta \mathrm{U}(\) Case-I \()\gt\Delta \mathrm{U}(\) Case-II \()\gt\Delta \mathrm{U}(\) Case-III \()\)
Answer & Solution
Correct Answer
(A) \(\Delta \mathrm{U}(\) Case-I \()=\Delta \mathrm{U}(\) Case-II \()=\Delta \mathrm{U}(\) Case-III \()\)
Step-by-step Solution
Detailed explanation
As internal energy ' \(U\) ' is a state function, its cyclic integral must be zero in a cyclic process \(\therefore \Delta U \text { case }(I)=\Delta U \text { case }(\text { II })=\Delta U \text { case }(\text { III })\)
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