Verified Solution • 2025–26 Session moderate

QJEEM19Ph5cca: JEE Main 2019 Physics - Class 11 JEEM19Ph5cca Solution: A rigid diatomic ideal gas undergoes an adiabatic process at…

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

Multiple Choice Options

  • A 53{5 \over 3}
  • B 25{2 \over 5}
  • C 35{3 \over 5}
  • D 23{2 \over 3}

Step-by-Step Verified Solution

For adiabatic process : TV γ\gamma −- 1 = constant

For diatomic process : γ\gamma −- 1 = 75−1{7 \over 5} - 1

∴\therefore   x = 25{2 \over 5}

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