In an adiabatic process, density of a diatomic gas becomes 32 times its initial value. final pressur

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11.Thermodynamics

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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

A

$326$

B

$\frac{1}{32}$

C

$32$

D

$128$

(JEE MAIN-2020)

Solution

In adiabatic process

$PV ^{\gamma}= constant$

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

as mass is constant

$P \propto \rho^{\gamma}$

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

Standard 11

Physics

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