To a large spherical container filled with gas, we add a fine tube of section A in which a steel ball of mass M can slide (Fig. 5.5). We wish to determine the γ coefficient of the gas. The ball is dropped in the tube and oscillates at a frequency f, which can easily be measured. This process is

Question

To a large spherical container filled with gas, we add a fine tube of section A in which a steel ball of mass M can slide (Fig. 5.5). We wish to determine the γ coefficient of the gas. The ball is dropped in the tube and oscillates at a frequency f, which can easily be measured. This process is assumed reversible. However, the measurement of the motion of the ball is sufficiently fast that the process can be considered adiabatic. We denote by [latex]V_0[/latex] the volume and [latex]p_0[/latex] the pressure at equilibrium and by [latex]p^{ext}[/latex] the external pressure, considered constant.

1. Determine to first-order the volume V (z) and the pressure p (z) as a function of the vertical displacement z of the ball in the limit of small displacements, i.e. [latex] \ll V_0[/latex].
2. The elastic force exerted by the gas on the ball is given by [latex]F = (p (z) − p^{ext]) A \hat{z}[/latex]. Determine the ball equation of motion.
3. Deduce from it the expression for the γ coefficient in terms of the frequency f of the oscillations around the equilibrium position, the pressure [latex]p_0[/latex] and the volume [latex]V_0[/latex] of the gas, and the cross-section A of the tube.

Accepted Answer

1. The volume of the gas V (z) is the sum of the volume [latex]V_0[/latex] at equilibrium and of the change in volume Az due to the displacement z of the ball, i.e.

[latex]V (z) = V_0 + Az[/latex].

As the process is isentropic (5.90), the pressure and the volume satisfy the following relation,

[latex]pV^γ = const[/latex].

[latex]p (z) V (z)^γ = p_0 V^γ_0[/latex].

Performing a first-order series expansion in terms of the small dimensionless term [latex]Az/V_0[/latex], the pressure is given by,

[latex]p (z) = p0\biggl(\frac{1}{1+\frac{A}{V_0}z } \biggr)^\gamma \simeq p_0\Bigl(1-\gamma \frac{A}{V_0}z \Bigr)[/latex].

2. Newton’s second law applied to the ball yields,

Mg + F = Ma

Projecting this equation on a vertical axis of unit vector [latex]\hat{z}[/latex] yields,

[latex]−Mg +(p (z) − p^{ext})A = M\ddot{z}[/latex].

This equation can be rewritten in the usual manner for a harmonic oscillator,

[latex]\ddot{z}+\frac{\gamma A^2p_0}{MV_0}z-\frac{1}{M}\Bigl((p_0-p^{ext})A-Mg\Bigr)=0[/latex].

3. The frequency f of the oscillations around the equilibrium position is given by,

[latex]f=\frac{\omega }{2\pi }=\frac{1}{2\pi \sqrt{\frac{\gamma A^2p_0}{MV_0} } }[/latex].

When solving this equation for γ, we obtain,

[latex]\gamma =\frac{4\pi ^2f^2MV_0}{A^2p_0}[/latex].

Question 5.9.5: To a large spherical container filled with gas, we add a fin...

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