Question 966: Thermodynamics & KTG

Question

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram.
If Q
1
, Q
2
, Q
3
indicate the heat absorbed by the gas along the three processes and $\Delta $U
1
, $\Delta $U
2
,  $\Delta $U
3
indicate the change in internal energy along the three processes respectively, then
![](https://neetcare-image.cdn.examgoal.net/1qpq2yck7l83w4k/9b7b2740-bfb4-4e9e-8712-9c50163fbd4e/82e29240-626f-11ea-bddc-5f310cab17d0/file-1qpq2yck7l83w4l-293w.jpg)

Accepted Answer

Initial and final condition is same for all process
$\Delta {U_1} = \Delta {U_2} = \Delta {U_3}$
from first law of thermodynamics
$\Delta Q = \Delta U + \Delta W$
Work done
$\Delta {W_1} > \Delta {W_2} > \Delta {W_3}$   (Area of P.V. graph)
So $\Delta {Q_1} > \Delta {Q_2} > \Delta {Q_3}$

Answer

Q
1
> Q
2
> Q
3
and  $\Delta $U
1
= $\Delta $U
2
= $\Delta $U
3

Answer

Q
3
> Q
2
> Q
1
and  $\Delta $U
1
= $\Delta $U
2
= $\Delta $U
3

Answer

Q
1
= Q
2
= Q
3
and  $\Delta $U
1
> $\Delta $U
2
> $\Delta $U
3

Answer

Q
3
= Q
2
= Q
1
and  $\Delta $U
1
> $\Delta $U
2
> $\Delta $U
3

Thermodynamics & KTG

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