How can adiabatic pressure variations be used to determine specific heat ratio?

[PhizKid]

1. Homework Statement

The following describes a method used to measure the specific heat ratio $\gamma = c_p/c_V$ of a gas. The gas, assumed ideal, is confined within a vertical cylindrical container and supports a freely moving piston of mass $m$. The piston and cylinder both have the same cross-sectional area $A$. Atmospheric pressure is $p_0$, and when the piston is in equilibrium under the influence of gravity (acceleration $g$) and the gas pressure, the volume of the gas is $V_0$. The piston is now displaced slightly from its equilibrium position and is found to oscillate about this position with frequency $\nu$. The oscillations of the piston are slow enough that the gas always remains in internal equilibrium, but fast enough that the gas cannot exchange heat with the outside. The variations in gas pressure and volume are thus adiabatic. Express $\gamma$ in terms of $m, g, A, p_0, V_0$, and $\nu$.

The Attempt at a Solution

Denote by $p$ the pressure exerted by the gas and by $V$ the volume below the piston. The gas undergoes a quasi-static adiabatic change so at any given instant during the piston's oscillation the relation $p V^{\gamma} = \text{const.}$ holds so this then implies that $(p(0) + \delta p)(V_0 + \delta V)^{\gamma} = p(0) V^{\gamma}_0$ i.e. $(p(0) + \delta p)(1 + \delta V/V_0)^{\gamma} = p(0)$. Now the oscillations about mechanical equilibrium are small so at any given instant during the oscillation, $\delta V \ll V_0, \delta p \ll p(0)$ hence $$(p(0) + \delta p)(1 + \delta V/V_0)^{\gamma} \\= (p(0) + \delta p)(1 + \gamma \delta V/V_0 + O((\delta V/V_0)^2) ) \\= p(0) + \delta p + \gamma p(0) \delta V/V_0 + O((\delta V/V_0)^2) = p(0)$$ therefore $$\gamma = -\frac{V_0}{p(0)}\frac{\delta p}{\delta V} \\= -\frac{V_0}{p(0)A}\frac{\delta (pA)}{\delta (A y)} \\= -\frac{V_0}{(mgA + p_0 A^2)}\frac{\delta F}{\delta y}$$ where $F$ is the force exerted by the gas on the piston at any given instant during the oscillation, so that $\delta F = F - F(0)$, and $\delta y$ is the displacement from the equilibrium position at that given instant. We also have that $F - p_0 A - mg = F_{\text{piston}} = -k \delta y = -\omega^2 m \delta y$ where I have used the fact that the piston undergoes simple harmonic motion (there is no damping since the gas loses no heat during the oscillations). Then $\delta F = (-\omega^2 m\delta y + p_0A + mg) - (p_0 A + mg) = -\omega^2 m \delta y$ so $\gamma = \frac{\omega^2 m V_0}{(mgA + p_0 A^2)} = \frac{4\pi^2 \nu^2 m V_0}{(mgA + p_0 A^2)}$ where I have used the fact that $\omega = 2\pi \nu$ for the angular frequency. This is what the book lists as the correct expression for $\gamma$ but I just wanted to make sure my solution was valid, particularly the steps taken within this very paragraph. Thanks in advance!

 

[Chestermiller]

I wasn't able to follow the details of what you did, but here is how I would do it.

Force Balance on Piston:

$$(p-p_0)A-mg=ma=\frac{m}{A}\frac{d^2V}{dt^2}$$
or
$$(\frac{k}{V^{γ}}-p_0)A-mg=\frac{m}{A}\frac{d^2V}{dt^2}$$

Initially,
$$(\frac{k}{V_0^{γ}}-p_0)A-mg=0$$

If we subtract the initial condition from the force balance we obtain:

$$k(\frac{1}{V^{γ}}-\frac{1}{V_0^{γ}})A=\frac{m}{A}\frac{d^2V}{dt^2}$$
If we write V = V₀+V', and linearize with respect to V', we obtain:
$$-\frac{kγ}{V_0^{{γ+1}}}AV'=\frac{m}{A}\frac{d^2V'}{dt^2}$$
From the initial condition:
$$\frac{k}{V_0^{γ}}=\frac{p_0A+mg}{A}$$
So,
$$-\frac{γ(p_0A+mg)}{V_0}V'=\frac{m}{A}\frac{d^2V'}{dt^2}$$
So:
$$ω^2=\frac{γ(p_0A+mg)A}{mV_0}$$

Chet

 

[PhizKid]

Thanks, but I wanted to know if my solution was correct. I didn't want someone to just tell me their own solution.

 

[Chestermiller]

Thanks, but I wanted to know if my solution was correct. I didn't want someone to just tell me their own solution.

Yes. Our solutions match. I thought you might find it interesting to see it done a different way.

Chet

 

[PhizKid]

Thank you! I see now that our solutions are, apart from a few notational differences and steps, basically the same. Is that what you meant when you said that our solutions matched?

 

[Chestermiller]

Thank you! I see now that our solutions are, apart from a few notational differences and steps, basically the same. Is that what you meant when you said that our solutions matched?

Yes.