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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2019
  • Medium
  • MCQ

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The rational between temperature and volume for the process is \(TV^x =\) constant, then \(x\) is

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Answer & Solution

Correct Answer

(B) \(2/5\)

Step-by-step Solution

Detailed explanation

Equation of adiabatic process

\(TV^{2/f} =\) constant

\(\therefore \,\,\frac {2}{f}=\frac {2}{5}=x\)

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