64. Concept Questions The drawing shows a gas confined to a cylinder by a massle

Question

64. Concept Questions The drawing shows a gas confined to a cylinder by a massless piston that is attached to an ideal spring. Outside the cylinder is a vacuum. The cross- sectional area of the piston is A. The initial R pressure, volume, and temperature of the gas are, respectively, Po, Vo, and To, and the spring is initially stretched by an amount o with respect to its unstrained length. The gas is heated, so that its final pressure, volume, and temperature are Pt, ½, and T, and the spring is stretched by an amount with re- spect to its unstrained length. (a) What is the relation between the magnitude of the force required to stretch an ideal spring and the amount of the stretch with respect to the.unstrained length of the spring? (b) What are the magnitudes of the forces that the initial and final pressures apply to the piston and, hence, to the spring? Express your answers in terms of the pressures and the cross-sec- tional area of the piston. (c) According to the ideal gas law, how are the initial pressure, volume, and temperature related to the final pressure, volume, and temperature? (d) How is the final volume re- lated to the initial volume, the cross-sectional area of the piston, and the initial and final amounts by which the spring is stretched Account for your answer. Problem The initial temperature and volume of the gas described in the Concept Questions are 273 K and 6.00 x 104 m2. The initial and final amounts by which the spring is stretched are, respectively, 0.0800 and 0.1000 m. The cross-sectional area of the piston is 2.50 × 10-3 m2, what is the final temperature of the gas? Piston

Explanation / Answer

a)F = k*x

Initial force by the gas on the piston = Po*A

so, balancing the forces,

Po*A = k*xo

b)initial force = Po*A

final force = Pf*A

c)Ideal gas law states that,

Po*Vo/To = Pf*Vf/Tf

d)Vf/A - Vo/A = xf - xo

so,

using the ideal gas equation,

Po*Vo/To = Pf*Vf/Tf,

xo*Vo/(To*A) = xf*Vf/(Tf*A)

or 0.08*6*10^-4/(273) = 0.1*V/(T)

also,

using the equation of d,

6*10^-4/(2.5*10^-3) - V/(2.5*10^-3) = 0.08-0.1

or V = 6.5*10^-4 m^3

T = 369.68 K

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