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

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:

1. 32

2. 326

3. 128

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

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

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

$\Rightarrow \frac{n P_{i}}{P_{i}}=\left(2^{5}\right)^{7 / 5}=2^{7}$

$\Rightarrow n=2^{7}=128$