Question

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

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

Answer 1

Answer: (C)

Here, $\gamma=\frac{7}{5}, P_{1}=P, \rho_{1}=\rho; P_{2}=P', \rho_{2}=\rho'$
For an adiabatic process, $PV^{\gamma} =$ constant
$\therefore P_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma}$
$\frac{P_{2}}{P_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(\frac{\rho_{2}}{\rho_{1}}\right)^{\gamma}=\left(\frac{\rho'}{\rho}\right)^{7/5}=\left(32\right)^{7/5}$
$\frac{P'}{P}=\left(2^{5}\right)^{7/5}=2^{7}=128$