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JEE Mains · Physics · STD 11 - 11. thermodynamics

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

  1. A \(326\)
  2. B \(\frac{1}{32}\)
  3. C \(32\)
  4. D \(128\)
Verified Solution

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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