Question 394: Thermodynamics & KTG

Question

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 :

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

$$\Delta $$Q
A
> $$\Delta $$Q
B
; $$\Delta $$U
A
> $$\Delta $$U
B

Answer

$$\Delta $$Q
A
< $$\Delta $$Q
B
; $$\Delta $$U
A
< $$\Delta $$U
B

Answer

$$\Delta $$Q
A
> $$\Delta $$Q
B
; $$\Delta $$U
A
= $$\Delta $$U
B

Answer

$$\Delta $$Q
A
= $$\Delta $$Q
B
; $$\Delta $$U
A
= $$\Delta $$U
B

Thermodynamics & KTG

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 :

Choose the correct answer: