Question

Consider an ideal gas enclosed in a vertical cylindrical container supported by a freely moving piston of mass $M$ . When the piston is in equilibrium, the volume of the gas is $V_{0}$ and its pressure is $P_{0}.$ The piston is slightly displaced from the equilibrium position and released. Frequency of SHM (simple harmonic motion) of piston is $\frac{1}{2 \pi }\sqrt{\frac{\gamma P_{0} A^{N}}{M^{q} V_{0}}}$ . What is the value of $\frac{2 q}{N}=?$

Answer 1

At equilibrium,
$P_{atm}A+MG=P_{0}A....\left(i\right)$
When piston is pushed down a distance $x$
$\left(P_{0} + d P\right)-\left(P_{\text{atm }} A + M g\right)=M\frac{d^{2} x}{d t^{2}}....\left(i i\right)$
For an adiabatic system,
$PV^{\gamma }=$ constant
$\Rightarrow V^{\gamma }dP+\gamma PV^{\gamma - 1}dV=0$
$\Rightarrow dP=\frac{- \gamma PdV}{V}=\frac{- \gamma P_{0} \left(\right. Ax \left.\right)}{V_{0}}....\left(i i i\right)$
From (i), (ii) \& (iii)
$M\frac{d^{2} x}{d t^{2}}=\frac{- \gamma P_{0} A^{2}}{V_{0}}x$
$\Rightarrow \frac{d^{2} x}{dt^{2}}=\frac{- \gamma P_{0} A^{2}}{V_{0} M}x$
$\Rightarrow \omega =\sqrt{\frac{\gamma P_{0} A^{2}}{V_{0} M}}$ and $v=\frac{1}{2 \pi }\sqrt{\frac{\gamma P_{0} A^{2}}{V_{0} M}}$
$\Rightarrow N=2;q=1$
Now $\frac{2 q}{N}=\frac{2}{2}=1$