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 $n$ is:
Correct Option: , 3
Solution:
(3) In adiabatic process
$P V^{\gamma}=$ constant
$\therefore P\left(\frac{m}{\rho}\right)^{\gamma}=$ constant $\left(\because V=\frac{m}{\rho}\right)$
As mass is constant
$\therefore P \propto \rho^{\gamma}$
If $P_{i}$ and $P_{f}$ be the initial and final pressure of the gas and $\rho_{i}$ and $\rho_{f}$ be the initial and final density of the gas. Then
$\frac{P_{f}}{P_{i}}=\left(\frac{\rho_{f}}{\rho_{i}}\right)^{\gamma}=(32)^{7 / 5}$
$\Rightarrow \frac{n P_{i}}{P_{i}}=\left(2^{5}\right)^{7 / 5}=2^{7}$
$\Rightarrow n=2^{7}=128$