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:

• A. $32$
• B. $326$
• C. $128$
• D. $\frac{1}{32}$

Answer 1

The correct option is C $128$

In adiabtaic process

$P{V}^{\gamma }=\text{constant}$
$\therefore P{\left(\frac{m}{\rho }\right)}^{\gamma }=\text{constant}(\because V=\frac{m}{\rho })$
As mass is constant

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

For diatomic gas $\gamma =7/5$

If $P_i$ and $P_f$ be the initial and final pressure of the gas and If ${\rho }_i$ and ${\rho }_f$ be the initial and final density of the gas. Then

$\frac{P_f}{P_i}={\left(\frac{{\rho }_f}{{\rho }_i}\right)}^{\gamma }=(32{)}^{7/5}$
$\Rightarrow \frac{n{P}_i}{P_i}=(2^5{)}^{7/5}=2^7$
$\Rightarrow n=2^7=128$