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NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice

When 500 calories heat is given to the gas X in an isobaric process, its work done comes out as 142 .8 calories. The gas X is

Options

  1. AO 2
  2. BN H 3
  3. CH e
  4. DS O 2

Correct answer

A. O 2

Step-by-step solution

∵ ΔT = W nR ∴   Q = nC p (ΔT) = nC p W nR = C p W R C p   =   QR W   =   500   ×   2 142 .8   =   7 C p   =   7,  indicates that the gas is diatomic. Thus, it should be O 2

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