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 $\frac{2q}{N}=?$

Answer 1

Answer: 1

At equilibrium,
$P_{atm}A+MG=P_{0}A....\left(i\right)$
When piston is pushed down a distance $x$
$\left(P_0+dP\right)-\left(P_{\text{atm }}A+Mg\right)=M\frac{d^{2}x}{d{t}^2}....\left(ii\right)$
For an adiabatic system,
$P{V}^{\gamma }=$ constant
$\Rightarrow V^{\gamma }dP+\gamma P{V}^{\gamma -1}dV=0$
$\Rightarrow dP=\frac{-\gamma PdV}{V}=\frac{-\gamma P_0(Ax)}{V_0}....\left(iii\right)$
From (i), (ii) \& (iii)
$M\frac{d^{2}x}{d{t}^2}=\frac{-\gamma P_0{A}^2}{V_0}x$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=\frac{-\gamma P_0{A}^2}{V_{0}M}x$
$\Rightarrow \omega =\sqrt{\frac{\gamma P_0{A}^2}{V_{0}M}}$ and $v=\frac{1}{2\pi }\sqrt{\frac{\gamma P_0{A}^2}{V_{0}M}}$
$\Rightarrow N=2;q=1$
Now $\frac{2q}{N}=\frac{2}{2}=1$