Q. The pressure and density of a diatomic gas $(\gamma = \frac{7}{5})$ change adiabatically from $\left( p_{1},\rho_{1} \right)to\ \left( p_{1},\rho_{2} \right).$ If $\frac{\rho_{1}}{\rho_{2}} = 32,\ then\frac{p_{1}}{p_{2}}$ should be

A) 16
B) 32
C) 64
D) 128
Explanation

(d) In an adiabatic process,
$\ \ \ \ \ \ \ \ \ \ \ \ pV^{\gamma} = constant$
or ${\ \ \ \ p}_{1}V_{1}^{\gamma} = p_{2}V_{2}^{\gamma}$
or $\frac{P_{1}}{P_{2}} = \left( \frac{V_{2}}{V_{1}} \right)^{\gamma}$ ----(i)
Volume of gas $= \frac{Mass}{Density}$
ie, $V = \frac{M}{\rho}\ \ \ or\ \ V \propto \frac{1}{\rho}$
$\ \ \ \ \ \ \therefore\frac{V_{1}}{V_{2}} = \frac{\rho_{2}}{\rho_{1}} = 32$
Thus, from Eq.(i), we have
$\frac{P_{2}}{P_{1}} = (32)^{\gamma} = (32)^{7/5} = 2^{7} = 128$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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