Question

The pressure and density of a diatomic gas $\left(\gamma =\frac{7}{5}\right)$ change adiabatically from $\left(P,\rho \right)$ to $\left(P\prime ,\rho \prime \right).$
If $\frac{{\rho }^{\prime }}{\rho }=32$, then $\frac{P^{\prime }}{P}$ is

• A. $\frac{1}{128}$
• B. $32$
• C. $128$
• D. $256$

Answer 1

Answer: (C)

Here, $\gamma =\frac{7}{5},P_1=P,{\rho }_1=\rho ;P_2=P^{\prime },{\rho }_2={\rho }^{\prime }$
For an adiabatic process, $P{V}^{\gamma }=$ constant
$\therefore P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
$\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }={\left(\frac{{\rho }^{\prime }}{\rho }\right)}^{7/5}={\left(32\right)}^{7/5}$
$\frac{P^{\prime }}{P}={\left(2^5\right)}^{7/5}=2^7=128$