• # question_answer Following figure shows two processes A and B for a gas. If $\Delta {{Q}_{A}}$and $\Delta {{Q}_{B}}$are the amount of heat absorbed by the system in two cases, and $\Delta {{U}_{A}}$and $\Delta {{U}_{B}}$are changes in internal energies, respectively, then : [JEE Main 9-4-2019 Morning] A) $\Delta {{Q}_{A}}=\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}$B) $\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}$C) $\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}>\Delta {{U}_{B}}$D) $\Delta {{Q}_{A}}<\Delta {{Q}_{B}};\Delta {{U}_{A}}<\Delta {{U}_{B}}$

Initial and final states for both the processes are same. $\therefore$$\Delta {{U}_{A}}=\Delta {{U}_{B}}$ Work done during process A is greater than in process B. By First Law of thermodynamics $\Delta Q=\Delta U+W$$\Rightarrow$$\Delta {{Q}_{A}}>\Delta {{Q}_{B}}$