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 :
![](https://imagex.cdn.examgoal.net/18hoxe66ijwhmt4d2/c172e2f6-d582-488e-b359-c16a3d7f959e/7c84af60-86af-11e9-8081-1d558b7f1d26/file-18hoxe66ijwhmt4d3.png)

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: