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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##.
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 [tex](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)[/tex] therefore [tex]\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} [/tex] 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!
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 [tex](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)[/tex] therefore [tex]\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} [/tex] 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!