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Thermodynamic process

An ideal gas enclosed in a vertical cylindrical container supp

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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
(a) $\frac{1}{2 π} \frac{A _{γ} p _{0}}{V _{0} M}$
(b) $\frac{1}{2 π} \frac{V _{0} M p _{0}}{A ^{2} γ}$
(c) $\frac{1}{2 π} \frac{A ^{2} γ p _{0}}{M V _{0}}$
(d) $\frac{1}{2 π} \frac{M V _{0}}{A _{γ} p _{0}}$

Views: 5,478 students

Updated on: May 13, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2013

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Evelyn Discussed

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
(a)

\frac{1}{2 π} \frac{A _{γ} p _{0}}{V _{0} M}

(b)

\frac{1}{2 π} \frac{V _{0} M p _{0}}{A ^{2} γ}

(c)

\frac{1}{2 π} \frac{A ^{2} γ p _{0}}{M V _{0}}

(d)

\frac{1}{2 π} \frac{M V _{0}}{A _{γ} p _{0}}

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Exp. (c)
Pressure applied by piston,

\frac{M g}{A} = p_{0}

As no exchange of heat, so process is adiabatic.

p_{0} V_{0}^{γ} = p V^{γ}, p_{0} A x_{0}^{γ} = p A (x_{0} - x)^{γ}

p = \frac{p _{0} x _{0}^{γ}}{( x _{0} - x ) ^{γ}}

Let piston be displaced by

x

M g - (\frac{p _{0} x _{0}^{γ}}{( x _{0} - x ) ^{γ}}) A = f_{res t or in g}

p_{0} A (1 - \frac{x _{0}^{γ}}{( x _{0} - x ) ^{γ}}) = F_{res t or in g} [∵ x_{0} - x \approx x_{0}]

F = - \frac{γ p _{0} A x}{x _{0}}

∴ f = \frac{1}{2 π} \frac{γ p _{0} A}{x _{0} M} = \frac{1}{2 π} \frac{γ p _{0} A ^{2}}{M V _{0}}

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Question Text	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 (a) $\frac{1}{2 π} \frac{A _{γ} p _{0}}{V _{0} M}$ (b) $\frac{1}{2 π} \frac{V _{0} M p _{0}}{A ^{2} γ}$ (c) $\frac{1}{2 π} \frac{A ^{2} γ p _{0}}{M V _{0}}$ (d) $\frac{1}{2 π} \frac{M V _{0}}{A _{γ} p _{0}}$
Updated On	May 13, 2023
Topic	Thermodynamics
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 8
Upvotes	932
Avg. Video Duration	7 min

Text solutionVerified

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