Question 452: Thermodynamics & KTG

Question

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 :

Accepted Answer

Initial and final states for both the processes are
same,
$$ 	herefore $$ $$\Delta $$U
A
= $$\Delta $$U
B
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
$$ \Rightarrow $$ $$\Delta $$Q
A
> $$\Delta $$Q
B

Answer

$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA > $$\Delta $$UB

Answer

$$\Delta $$QA < $$\Delta $$QB ; $$\Delta $$UA < $$\Delta $$UB

Answer

$$\Delta $$QA > $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB

Answer

$$\Delta $$QA = $$\Delta $$QB ; $$\Delta $$UA = $$\Delta $$UB

Thermodynamics & KTG

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 :

Choose the correct answer: