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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2006 · Physics · Thermodynamics

Match the following for the given process

Match the following for the given process Match the following for the given process
Match the following for the given process

Options

  1. A(a) s, (b) p, r, (c) r, (d) q,s
  2. B(a) r, (b) p, s, (c) q, (d) q
  3. C(a) s, (b) p, r, (c) q, (d) q
  4. D(a) r, (b) p, s, (c) r, (d) q,s

Correct answer

A. (a) s, (b) p, r, (c) r, (d) q,s

Step-by-step solution

(a) →s\rightarrow \mathrm{s} (b) →p,r\rightarrow \mathrm{p}, \mathrm{r}, (c) →r\rightarrow \mathrm{r} (d) →q,s\rightarrow \mathrm{q}, \mathrm{s} In process J→K:V\mathbf{J} \rightarrow \mathbf{K}: V is constant whereas PP is decreasing, Therefore, TT should also decrease. \therefore \quad W=0, Δ\DeltaU=- ve and \text{ ve and }Q < 0 In process K→L:P\mathbf{K} \rightarrow \mathbf{L}: P is constant while VV is increasing. Therefore, temperature should also increase. \therefore \quad W>0, Δ\DeltaU>0  and \text{ and }Q>0 In process L→M\mathbf{L} \rightarrow \mathbf{M} : This is inverse of process J→KJ \rightarrow K. \therefore \quad W=0, Δ\DeltaU>0  and \text{ and }Q>0 In process M→J:V\mathbf{M} \rightarrow \mathbf{J}: V is decreasing. Therefore, W<0W \lt 0(P V)J<(PV)M∴TJ<TM or \DeltaU<0 Therefore, Q<0.\begin{array}{lcl}& (P \mathrm{~V})_J & < (P V)_M \\ \therefore & T_J & < T_M \text{ or } \quad \DeltaU < 0 \\ \text{ Therefore, }& Q & < 0 . \end{array}No,solutionisrequired.No, solution is required.
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