CBSE-XI-Physics
26: Laws of Thermodynamics
- #3Figure shows two processes A and B on a system. Let â–µQ1 and â–µQ2 be the heat given to the system in processes A and B respectively. Then
(a) â–µQ1 > â–µQ2
(b) â–µQ1 = â–µQ2
(c) â–µQ1 < â–µQ2
(d) ▵Q1 ≤ ▵Q2.
FiguredigAnsr: aAns :
(a) ∆Q1 > ∆Q2
Both the processes A and B have common initial and final points. So, change in internal energy, ∆U is same in both the cases. Internal energy is a state function that does not depend on the path followed.
In the P-V diagram, the area under the curve represents the work done on the system, ∆W. Since area under curve A > area under curve B, ∆W​1> ∆W2.
Now,
`` \Delta {Q}_{1}=\Delta U+\Delta {W}_{1}``
`` \Delta {Q}_{2}=\Delta U+\Delta {W}_{2}``
`` \,\mathrm{\,But\,}\Delta {W}_{1}>\Delta {W}_{2}``
`` \Rightarrow \Delta {Q}_{1}>\Delta {Q}_{2}``
Here, ∆Q1 and ∆Q2 denote the heat given to the system in processes A and B, respectively.
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