Question

A piston with a radius of 2.5 cm is placed into a cylindrical tube a distance of 4.5 cm from the closed end of the tube, holding in a quantity of air. The piston is attached to a spring with a spring constant of 324 N/m. The other end of the spring is attached to a rigid wall. The tube is surrounded in ice water which keeps the gas at a temperature of o°C. Initially, the spring is relaxed, and the gas under the piston has an absolute pressure of 2.25 105 Pa, as shown in the leftmost figure below. The gas then expands isothermally until the pressure drops such that the sum of the forces on the piston (the force of the spring plus the force due to the pressure of the gas) is equal to zero, as shown in the rightmost figure below. Ignore effects of gravity, assume a perfect vacuum above the top surface of the piston. (a) What is the initial volume of the gas? (b) How many moles of molecules are there in the gas? Hint: You know the initial pressure and temperature, and you found the initial volume in part (a). (c) How far does the gas compress the spring in the process? Hint: Set the forces to be equal, and then use the ideal gas law to relate the final gas pressure to the distance that the spring is compressed. Another hint: The final volume is just the final height of the piston times the piston area, and the final height of the piston is just the initial height plus the distance that the spring has been compressed. (d) Using the equation W = PdV , find the work done by the gas. Hint: PV = nRT. And

Answer 1

{a Radius of piston, I1= 2.5em = 2.5*10-2m area of piston, Aa To 12 = 22 x 2.5x102) ² an A=19.64 x 10-4 m2 Distance from closed and d = 4.5cm - 4.5x102m - Volume : Axd = 19.64 x 10-4 x 4.5 x 152. Tu= 88.38 x 10-6 m3 (b) initial voluone Via Va 88.38 ko bom? initial pressure, Pi: 2.25 x 105 Pa initial temperature, Ti = 0°C a 273 kr How using ideal gas equation i na pu = 2.25x105 x 8.8.38 x to 6 RT 8.3 x 273 A n=0.087759616-1 = 2.77x10 moles (c) Suppose spring is compressed by 'x'mete. total distance of piston from closed end adta a 4.5 x 10-2 + x on final volume = Ald+n) - 19.64*10-4 (d+x) . Pue nRT If The Constant than P. v. & P.₂ V ₂ Pi u = Pf vf Pha Pivi = 2.25x105x 88-38x106 uf 19.64610-4. (d4x) fga PXA ton due to gas - 2.25x105 x 8 8. B 8 x 106 x A +9.64*10-4 lata) I Eg = 2.25x105 8 8.38 x 106 x 19.64x1024 19.64610-4 I den) = 2. 2гу о хee-e x106 do tn) force due to spring, liven, ke 324 Non .. 11. fsz 3.242 Now, piston is in equilibrium. :. Eg Efs 2.25 x 105x88.38x100 a 3248. 0.06 = xlo. 045dn, x = 0.22 35 m Td) Walp.dv - (ORI du Puanda .: PanRa Wa MRT douf) = 8.77x10 3 8.5x273 x In Aulden). laxd 0.045+ 0.2235 0.045 Wa8.77x10-1 & 323 Tw=85.59