Verified Solution • 2025–26 Session moderate

QJEEM20Ph9ebd: JEE Main 2020 Physics - Class 11 JEEM20Ph9ebd Solution: In an adiabatic process, the density of a diatomic gas becomes…

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is :

Multiple Choice Options

  • A 128
  • B 32
  • C 326
  • D 132{1 \over {32}}

Step-by-Step Verified Solution

In adiabatic process

PV γ\gamma = constant

⇒\Rightarrow P(mρ)γP{\left( {{m \over \rho }} \right)^\gamma } = constant

As mass is constant

∴\therefore P ∝\propto ργ{{\rho ^\gamma }}

⇒\Rightarrow PfPi=(ρfρi)γ{{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \over {{\rho _i}}}} \right)^\gamma } = (32)75{\left( {32} \right)^{{7 \over 5}}} = 27 = 128

[ For diatomic gas γ\gamma = 75{{7 \over 5}} ]

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