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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. - Physics

MCQ
Fill in the Blanks

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, ∆W​1> ∆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.

Concept: Heat, Internal Energy and Work
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 4 Laws of Thermodynamics
MCQ | Q 3 | Page 61
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