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Question
Figure shows two processes A and B on a system. Let ∆Q1 and ∆Q2 be the heat given to the system in processes A and B respectively. Then ____________ .
Options
∆Q1 > ∆Q2
∆Q1 = ∆Q2
∆Q1 < ∆Q2
∆Q1 ≤ ∆Q2
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Solution
∆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, ∆W1> ∆W2.
Now,
∆Q1 = ∆U + ∆W1
∆Q2 = ∆U + ∆W2
But ∆W1 > ∆W2
⇒ ∆Q1 > ∆Q2
Here, ∆Q1 and ∆Q2 denote the heat given to the system in processes A and B, respectively.
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