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Thermodynamics question

2013 · Shift 2 · Q16
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Thermodynamics question

2013 · Shift 2 · Q16

JEE AdvancedChemistryThermodynamicsMCQ+3 / −1
A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure. JEE Advanced 2013 Paper 2 Offline Chemistry - Thermodynamics Question 17 English ComprehensionThe pair of isochoric processes among the transformation of states is
  1. A
    K to L and L to M
  2. B
    L to M and N to K
  3. C
    L to M and M to N
  4. D
    M to N and N to K
View written solutionFree

Correct answer: B

  1. Key idea: identify isochoric processes

    An isochoric process is one in which volume remains constant.

    On a thermodynamic graph:

    • On a PPP–VVV graph, isochoric processes are vertical lines.
    • On a VVV–TTT graph, isochoric processes are horizontal lines.
    • On a PPP–TTT graph, isochoric processes are straight lines through the origin since for an ideal gas, P=nRVTP = \frac{nR}{V}TP=VnR​T at constant VVV.

  2. From the given cyclic path K→L→M→N→KK \to L \to M \to N \to KK→L→M→N→K

    The two segments that represent constant volume are the ones where the volume does not change between the end states. From the figure, these are:

    • L→ML \to ML→M
    • N→KN \to KN→K

  3. Check the options

    • A: K→LK \to LK→L and L→ML \to ML→M
      Not correct.
    • B: L→ML \to ML→M and N→KN \to KN→K
      Correct.
    • C: L→ML \to ML→M and M→NM \to NM→N
      Not correct.
    • D: M→NM \to NM→N and N→KN \to KN→K
      Not correct.

  4. Final answer

    The pair of isochoric processes is: L→M and N→K\boxed{L \to M \text{ and } N \to K}L→M and N→K​

    So the correct option is B.

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