Question

The pressure and density of a diatomic gas $(\gamma =\frac{7}{5})$ changes from $(P,\rho )$ to $(P^{\prime },{\rho }^{\prime })$ during an adiabatic change. If $\frac{{\rho }^{\prime }}{\rho }=32$, then the value of $\frac{P^{\prime }}{P}$ is

• A. $36$
• B. $\frac{1}{128}$
• C. $128$
• D. $\frac{1}{36}$

Answer 1

The correct option is C $128$

Given,
Initial state of gas $(P,\rho )$
Final state of gas $(P^{\prime },{\rho }^{\prime })$
We know that, mass density $\left(\rho \right)\propto \frac{1}{V}$
Thus, $\frac{V_1}{V_2}=\frac{{\rho }_2}{{\rho }_1}.....\left(1\right)$
where $V$ is the volume and $\rho$ is mass density.

Assuming no of moles of diatomic gas is constant.
From the equation of state of gas during an adiabatic process
$P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }$
we get, $\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }$,
Using equation $\left(1\right)$ and from the data given in the question,
$\frac{P^{\prime }}{P}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }={\left(\frac{{\rho }^{\prime }}{\rho }\right)}^{\gamma }=(32{)}^{\frac{7}{5}}$
$\Rightarrow \frac{P^{\prime }}{P}=2^7=128$
Thus, option (c) is the correct answer.