The pressure and density of a diatomic gas $(\gamma = 7/5)$ change from $(p,\ \rho)$ to $(p^{-},\ \rho^{-})$ during an adiabatic change. If $\frac{\rho^{'}}{\rho} = 32$, value of $\frac{p^{'}}{p}$ is

Question

Dr. Abubacker · Accepted Answer

(c) From $p_{1}V_{1}^{\gamma} = p_{1}V_{2}^{\gamma}$
$\frac{p_{2}}{p_{1}} = \left( \frac{V_{1}}{V_{2}} ight)^{\gamma} = \left( \frac{ho_{2}}{ho_{1}} ight)^{\gamma}$
$= (32)^{7/5} = \left( 2^{5} ight)^{7/5} = 2^{7} = 128$
$	herefore\ \frac{P_{2}}{P_{1}} = \frac{P^{\ }}{P} = 128$

Q. The pressure and density of a diatomic gas $(\gamma = 7/5)$ change from $(p,\ \rho)$ to $(p^{-},\ \rho^{-})$ during an adiabatic change. If $\frac{\rho^{'}}{\rho} = 32$, value of $\frac{p^{'}}{p}$ is

Explanation

Reviewed by Dr. Abubacker

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