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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2020
  • Medium
  • Numerical

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

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Answer & Solution

Correct Answer

(D) \(128\)

Step-by-step Solution

Detailed explanation

In adiabatic process

\(PV ^{\gamma}= constant\)

\(P \left(\frac{ m }{\rho}\right)^{\gamma}= constant\)

as mass is constant

\(P \propto \rho^{\gamma}\)

\(\frac{P_{f}}{P_{i}}=\left(\frac{\rho_{f}}{\rho_{i}}\right)^{\gamma}=(32)^{7 / 5}=2^{7}=128\)

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