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JEE Main201911 Jan 2019Morning ShiftPhysicsThermodynamicsActual

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV ^ x = constant, then x is:

Options

  1. A3 5
  2. B2 5
  3. C2 3
  4. D5 3

Correct answer

B. 2 5

Step-by-step solution

Equation of adiabatic change is TV ^ -1 = constant Put = 7 5 , we get: -1= 7 5 -1 x = 2 5

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