Question 6.TC.1A: Spring Cylinder With Massive Piston Consider n = 2 moles of ......
Spring Cylinder With Massive Piston
Consider n = 2 moles of ideal Helium gas at a pressure P_{0}, volume V_{0} and temperature T_{0} = 300 K placed in a vertical cylindrical container (see Fig. 6 – 1 ). A moveable frictionless horizontal piston of mass m = 10 kg (assume g =9. 8 m/s²) and cross sectionA = 500 cm² compresses the gas leaving the upper section of the container void. There is a vertical spring attached to the piston and the upper wall of the container. Disregard any gas leakage through their surface contact, and neglect the specific thermal capacities of the container, piston and spring. Initially the system is in equilibrium and the spring is unstretched. Neglect the spring’s mass.
(a) Calculate the frequency f of small oscillation of the piston, when it is slightly displaced from equilibrium position.
(b) Then the piston is pushed down until the gas volume is halved, and released with zero velocity. Calculate the value(s) of the gas volume when the piston speed is{\sqrt{\frac{4g V_{0}}{5A}}}\,.
Let the spring constantk={\frac{m g A}{V_{\mathrm{0}}}}.All the processes in gas are adiabatic.
Gas constant
R\;=\;8.314\;\mathrm{JK^{-1}m o l^{-1}}.• For mono-atomic gas (Helium) use 5 Laplace constant\gamma={\frac{5}{3}}.
(a) Gas Volume
At the initial condition, the system is in equilibrium and the spring is unstreched; therefore
P_{0}A=m g\quad\mathrm{or}\quad P_{0}=\frac{mg}{A}. (1)
The initial volume of gas
V_{\mathrm{0}}={\frac{n R T_{\mathrm{0}}}{P_{\mathrm{0}}}}={\frac{n R T_{\mathrm{0}}A}{m g}}. (2)
The work done by the gas from{\frac{1}{2}}\ V_{\mathrm{0}}to V
W_{\mathrm{gas}}=\int_{\frac{V_{0}}{2}}^{\mathrm{V}}{ P}d V=\int_{\frac{V_{0}}{2}}^{\mathrm{V}}{\frac{P_{0}V_{0}^{\gamma}}{V^{\gamma}}}d V={\frac{P_{0}V_{0}^{\gamma}}{1-\gamma}}\biggl[V^{1-\gamma}-\biggl({\frac{V_{0}}{2}}\biggr)^{1-\gamma}\biggr] (3)
Eq. (3) can also be obtained by calculating the internal energy change (without integration)
W_{\mathrm{gas}}=-\,\Delta E=-\,n C_{V}(\,T-\,T\,^{\prime}{_{0}})\,, (4)
where T_{0}^{\prime} is the temperature when the gas volume is{\frac{V_{0}}{2 }}.•
The change of the gravitational potential energy
\Delta E_{p}=m g\,\Delta h=m g\,\frac{V-\frac{1}{2}V_{o}}{A}. (5)
The change of the potential energy of the spring
\Delta_{\mathrm{spring}}={\frac{1}{2}}k x^{2}-{\frac{1}{2}}k x_{0}^{2}=\frac{1}{2}\left(\frac{m g A}{V_{\scriptscriptstyle0}}\right)\left(\frac{V_{\scriptscriptstyle0}-V}{A}\right)^{2}-\frac{1}{2}\left(\frac{m g A}{V_{\scriptscriptstyle0}}\right)\left[\frac{V_{\scriptscriptstyle0}-\frac{V_{\scriptscriptstyle0}}{2}}{A}\right]^{2}
=\frac{1}{2}\,\frac{m g V_{0}}{A}\Big(1-\frac{V}{V_{\mathrm{0}}}\Big)^{2}-\frac{1}{8}\Big(\frac{m g V_{0}}{A}\Big). (6)
The kinetic energy
E_{\mathrm{k}}={\frac{1}{2}}m\upsilon^{2}={\frac{1}{2}}m{\frac{4g V_{\mathrm{0}}}{5A}}={\frac{2m g V_{\mathrm{0}}}{5A}}. (7)
By conservation of energy, we have
W_{\mathrm{gas}}=\Delta E_{p}+\Delta_{\mathrm{spring}}+E_{\mathrm{k}}\,, (8)
\frac{P_{0}V_{0}^{\gamma}}{1-\gamma}\Big[V^{1-\gamma}-\Big(\frac{V_{0}}{2}\Big)^{1-\gamma}\Big]=m g\frac{V-\frac{V_{0}}{2}}{ A}+\frac{1}{2}\ \frac{m g V_{0}}{ A}\Big(1-\frac{V}{V_{0}}\Big)^{2}
-\,{\frac{1}{8}}\,{\frac{m g\,V_{0}}{A}}+{\frac{2}{5}}\,{\frac{m g\,V_{0}}{A}}\,, (9)
\frac{m g V_{0}}{A(1-\gamma)}\biggl[\frac{V^{1-\gamma}}{V_{0}^{1-\gamma}}-\biggl(\frac{1}{2}\biggr)^{1-\gamma}\biggr]=m g\,\frac{V-\frac{V_{0}}{2}}{A}+\frac{mgV_{0}}{2A}\biggl(1-\frac{V}{V_{0}}\biggr)^{2}
+\frac{11}{40}\frac{m g V_{0}}{A}. (10)
Let s={\frac{V}{V_{0}}}the above equation becomes
{\frac{1}{1-\gamma}}{\Big[}s^{1-\gamma}-{\Big(}{\frac{1}{2}}{\Big)}^{1-\gamma}{\Big]}={\Big(}s-{\frac{1}{2}}{\Big)}+{\frac{1}{2}}(1-s)^{2}+{\frac{11}{40}}. (11)
With \gamma={\frac{5}{3}}we get
0={\frac{1}{2}}s^{2}+{\frac{11}{40}}+{\frac{3}{2}}\Bigl[s^{-\frac{2}{3}}-\Bigl({\frac{1}{2}}\Bigr)^{-{\frac{2}{3}}}\Bigr]. (12)
Solving Eq. (12) numerically, we get
s_{1}=0.74{\mathrm{~and~}}s_{2}=1.30.Therefore V_{1}\;=\;0.74V_{0}\;=\;0.74\;\frac{n R T_{\scriptscriptstyle0}A}{m g}\;=\;1.88\;\mathrm{m}^{3}or \;V_{2}\;=\;1.30V_{0}\;=\; 3.31 m³•
(b) Small Oscillation
The equation of motion when the piston is displaced by x from the equilibrium position is
m\,x=-\,k x-P A+mg\,, (13)
P is the gas pressure
P={\frac{P_{\mathrm{0}}V_{\mathrm{0}}^{\gamma}}{V^{\gamma}}}={\frac{P_{\mathrm{0}}V_{\mathrm{0}}^{\gamma}}{(V_{\mathrm{0}}-A x)^{\gamma}}}
= \frac{P_{0}}{\left(1-\frac{A x}{V_{0}}\right)^{\gamma}}. (14)
Since Ax «Vo then we have P\approx P_{\mathrm{0}}\left(1+\gamma\frac{A x}{V_{\mathrm{0}}}\right), therefore
m x\approx {-}k x-P_{\mathrm{0}}A\Bigl(1+\gamma\frac{A x}{V_{\mathrm{0}}}\Bigr)+m g\,,
m x\approx-\Bigl[k+P_{0}A\Bigl(\gamma\frac{A}{V_{0}}\Bigr)\Bigr]x, (15)
m\,x\approx-\left[\frac{m g A}{V_{0}}+\frac{m g}{A}A\left(\gamma\frac{A}{V_{0}}\right)\right]x,
m x+(1+\gamma)\ \frac{m g A}{V_{0}}x\approx0.
The frequency of the small oscillation is
f={\frac{1}{2\pi}}{\sqrt{(1+\gamma)\ {\frac{g A}{V_{\mathrm{0}}}}}}
={\frac{1}{2\pi}}{\sqrt{(1+\gamma)\ {\frac{m g^{2}}{n R T_{\mathrm{0}}}}}}. (16)
Numerically f = 0.114 Hz.