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JEE Main 2013 (9 Apr, Shift 1) · P-V Diagrams & Cyclic Processes

P-V Diagrams & Cyclic Processes question from JEE Main 2013 (9 Apr, Shift 1)

Q77JEE Main 2013 · 9 APR · SHIFT 1Medium

P-V Diagrams & Cyclic Processes

Figure shows the variation in temperature (ΔT\Delta T) with the amount of heat supplied (Q) in an isobaric process corresponding to a monoatomic (M), diatomic (D) and a polyatomic (P) gas. The initial state of all the gases are the same and the scales for the two axes coincide. Ignoring vibrational degrees of freedom, the lines a, b and c respectively correspond to:

Figure

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Answer: (c)

  1. Isobaric: Q=nCpΔTQ = nC_p\Delta T, so slope of the QQ-ΔT\Delta T line is nCpnC_p.
  2. CpC_p: monoatomic 52R\frac{5}{2}R, diatomic 72R\frac{7}{2}R, polyatomic 4R4R.
  3. Steepest line a is polyatomic, b is diatomic, least steep c is monoatomic.
  4. So the answer is (c) P, D and M.
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