In an adiabatic process, the density of a diatomic gas

Question:

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 $\mathrm{n}$ is:

  1. 326

  2. $\frac{1}{32}$

  3. 32

  4. 128


Correct Option: , 4

Solution:

In adiabatic process

$\mathrm{PV} \gamma=$ constant

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

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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