Heat and Thermodynamics 6 Question 4

8. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and the cylinder have equal cross-sectional area $A$. 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. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

(2013 Main)

(a) $\frac{1}{2 \pi} \frac{A _{\gamma} p _0}{V _0 M}$

(b) $\frac{1}{2 \pi} \frac{V _0 M p _0}{A^{2} \gamma}$

(c) $\frac{1}{2 \pi} \sqrt{\frac{A^{2} \gamma p _0}{M V _0}}$

(d) $\frac{1}{2 \pi} \sqrt{\frac{M V _0}{A _{\gamma} p _0}}$

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Answer:

Correct Answer: 8. (c)

Solution:

  1. In equilibrium,

$$ p _0 A=M g $$

when slightly displaced downwards,

$d p=-\gamma\left(\frac{p _0}{V _0}\right) d V\left(\right.$ As in adiabatic process, $\left.\frac{d p}{d V}=-\gamma \frac{p}{V}\right)$

$\therefore$ Restoring force,

$$ \begin{aligned} F & =(d p) A=-\left(\frac{\gamma p _0}{V _0}\right)(A)(A x) \\ F & \propto-x \end{aligned} $$

Therefore, motion is simple harmonic comparing with

$$ \begin{aligned} F & =-k x \text { we have } \\ k & =\frac{\gamma p _0 A^{2}}{V _0} \\ \therefore \quad f & =\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{\gamma p _0 A^{2}}{M V _0}} \end{aligned} $$



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