JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 25)
Following figure shows two processes A and
B for a gas. If $$\Delta $$QA and $$\Delta $$QB are the amount of
heat absorbed by the system in two cases, and
$$\Delta $$UA and $$\Delta $$UB are changes in internal energies,
respectively, then :
_9th_April_Morning_Slot_en_25_1.png)
$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA > $$\Delta $$UB
$$\Delta $$QA < $$\Delta $$QB ; $$\Delta $$UA < $$\Delta $$UB
$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB
$$\Delta $$QA = $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB
Explanation
Initial and final states for both the processes are
same,
$$ \therefore $$ $$\Delta $$UA = $$\Delta $$UB
Work done during process A is greater than in process B. Because area is more By First law of thermodynamics
$$\Delta $$Q = $$\Delta $$U + W
$$ \Rightarrow $$ $$\Delta $$QA > $$\Delta $$QB
$$ \therefore $$ $$\Delta $$UA = $$\Delta $$UB
Work done during process A is greater than in process B. Because area is more By First law of thermodynamics
$$\Delta $$Q = $$\Delta $$U + W
$$ \Rightarrow $$ $$\Delta $$QA > $$\Delta $$QB
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